Bourbaki

3. Correspondences

1. Graphs and correspondences

Definition 1 03PG

G is said to be a graph if every element of G *is an ordered pair, i.e., if the relation

(∀z)(z∈G⇒(z is an ordered pair))(\forall z)(z \in G \Rightarrow (z \text{ is an ordered pair}))

is true.

If G is a graph, the relation (x,y)∈G(x, y) \in G is expressed by saying that "yy corresponds to xx under G".

Let R{x,y}\boldsymbol{R}\{x, y\} be a relation, where xx and yy are distinct letters. Let G\boldsymbol{G} be a letter, distinct from xx and yy, which does not appear in R\boldsymbol{R}. If the relation

(∃G)(G is a graph and (∀x)(∀y)(R⇔((x,y)∈G)))(\exists \boldsymbol{G})(\boldsymbol{G} \text{ is a graph and } (\forall x)(\forall y)(\boldsymbol{R} \Leftrightarrow ((x, y) \in \boldsymbol{G})))

is true, the relation R\boldsymbol{R} is said to have a graph (with respect to the letters xx and yy). The graph G\boldsymbol{G} is then unique by virtue of the axiom of extent, and is called the graph of R\boldsymbol{R} with respect to xx and yy.

¶ Let ZZ be a letter, distinct from xx and yy, which does not appear in RR. If the relation

(∃Z)(∀x)(∀y)(R⇒((x,y)∈Z))(\exists Z)(\forall x)(\forall y)(R \Rightarrow ((x, y) \in Z))

is true, then RR has a graph; we may take this graph to be the set of ordered pairs zz such that z∈Zz \in Z and R{pr1z, pr2z}R \{ \mathrm{pr}_1 z,\ \mathrm{pr}_2 z \} (zz being a letter distinct from xx, yy, ZZ which does not appear in RR). This condition is satisfied if we know a term TT, which does not contain either xx or yy, such that R⇒((x, y)∈T)R \Rightarrow ((x,\ y) \in T) is true.

Proposition 1 03HI

Let* G be a graph. There exists exactly one set A and exactly one set B with the following properties :

Cites 2

Cites II § 1. Collectivizing relations, Proposition 2, II § 2

(1) the relation (∃y)((x, y)∈G)(\exists y)((x,\ y) \in \mathrm{G}) is equivalent to x∈Ax \in \mathrm{A};

(2) the relation (∃x)((x, y)∈G)(\exists x)((x,\ y) \in \mathrm{G}) is equivalent to y∈By \in \mathrm{B}.

For it is sufficient to take A (resp. B) to be the set of all objects of the form pr1z\mathrm{pr}_1 z (resp. pr2z\mathrm{pr}_2 z), where z∈Gz \in \mathrm{G} (§ 1, no. 6). Precisely,

A=Ex((∃y)((x,y)∈G)))andB=Ey((∃x)((x, y)∈G));\mathrm{A} = \mathscr{E}_x((\exists y)((x, y) \in \mathrm{G}))) \qquad \text{and} \qquad \mathrm{B} = \mathscr{E}_y((\exists x)((x,\ y) \in \mathrm{G}));

these sets are called respectively the first and second projections of the graph G, or the domain and range of G; they are denoted by pr1⟨G⟩\mathrm{pr}_1 \langle \mathrm{G} \rangle and pr2⟨G⟩\mathrm{pr}_2 \langle \mathrm{G} \rangle (or by pr1G\mathrm{pr}_1 \mathrm{G} and pr2G\mathrm{pr}_2 \mathrm{G} if there is no risk of confusion). It is immediately verified that G⊂(pr1G)×(pr2G)\mathrm{G} \subset (\mathrm{pr}_1 \mathrm{G}) \times (\mathrm{pr}_2 \mathrm{G}); every set of ordered pairs is therefore a subset of a product, and conversely. If one of the two sets pr1G\mathrm{pr}_1 \mathrm{G}, pr2G\mathrm{pr}_2 \mathrm{G} is empty, we have G=∅\mathrm{G} = \emptyset (§ 2, Proposition 2).

Remark 03S1

The relation x=yx = y has no graph, for the first projection of the graph, if it existed, would be the set of all objects (cf. § 1, no. 7, Remark).

Cites 1

Cites II § 1. Collectivizing relations

Definition 2 03HJ

A correspondence between a set* A and a set B *is a triple

Γ=(G, A, B),\Gamma = (\mathrm{G},\ \mathrm{A},\ \mathrm{B}),

where G is a graph such that pr1G⊂A\mathrm{pr}_1 \mathrm{G} \subset \mathrm{A} and pr2G⊂B\mathrm{pr}_2 \mathrm{G} \subset \mathrm{B}. G is said to be the graph of Γ\Gamma, A is the source, and B the target of Γ\Gamma.

If (x, y)∈G(x,\ y) \in \mathrm{G}, we say that "yy corresponds to xx in the correspondence Γ\Gamma". For each x∈pr1Gx \in \mathrm{pr}_1 \mathrm{G} the correspondence Γ\Gamma is said to be defined at xx, and pr1G\mathrm{pr}_1 \mathrm{G} is called the domain of Γ\Gamma; each y∈pr2Gy \in \mathrm{pr}_2 \mathrm{G} is said to be a value taken by Γ\Gamma, and pr2G\mathrm{pr}_2 \mathrm{G} is called the range of Γ\Gamma.

If R{x,y}R \{ x, y \} is a relation which has a graph GG (with respect to the letters xx, yy), and if AA, BB are two sets such that pr1G⊂A\mathrm{pr}_1 G \subset A and pr2G⊂B\mathrm{pr}_2 G \subset B, we say that RR is a relation between an element of AA and an element of BB (with respect to the letters xx, yy). The correspondence Γ=(G,A,B)\Gamma = (G, A, B) is said to be the correspondence between AA and BB defined by the relation RR (with respect to xx and yy).

Let G be a graph and let X be a set. The relation

x∈X and (x,y)∈Gx \in X \text{ and } (x, y) \in G

implies (x,y)∈G(x, y) \in G and therefore has a graph G′G'. The second projection of G′G' consists of all the objects which correspond under G to objects of X.

Definition 3 03PI

Let* G be a graph and X a set. The set of all objects which correspond under G to elements of X is called the image of X under G and is denoted by G⟨X⟩G\langle X \rangle or G(X)G(X).

Let Γ=(G,A,B)\Gamma = (G, A, B) be a correspondence and let X be a subset of A. The set G⟨X⟩G\langle X \rangle is also denoted by Γ⟨X⟩\Gamma\langle X \rangle or Γ(X)\Gamma(X) and is called the image of X under Γ\Gamma.

Remarks

Cited by 1

Cited by Remark

Remark 1 03PH

Precisely, G⟨X⟩G\langle X \rangle denotes the set Ey((∃x)(x∈X and (x,y)∈G))\mathscr{E}_y((\exists x)(x \in X \text{ and } (x,y) \in G)). From now on we shall not usually translate our definitions into formal language.

Remark 2 03S2

The notations G(X)G(X) and Γ(X)\Gamma(X) can occasionally lead to confusion with the notation introduced later (cf. no. 4, Remark following Definition 9).

Let G be a graph. Since the relation (x,y)∈G(x, y) \in G implies y∈pr2Gy \in \mathrm{pr}_2 G, we have G⟨X⟩⊂pr2GG\langle X \rangle \subset \mathrm{pr}_2 G for every set X. Since (x,y)∈G(x, y) \in G implies x∈pr1Gx \in \mathrm{pr}_1 G, we have G⟨pr1G⟩=pr2GG\langle \mathrm{pr}_1 G \rangle = \mathrm{pr}_2 G. We have G⟨∅⟩=∅G\langle \emptyset \rangle = \emptyset, since x∉∅x \notin \emptyset is a theorem. If X⊂pr1GX \subset \mathrm{pr}_1 G and X≠∅X \neq \emptyset, we have G⟨X⟩≠∅G\langle X \rangle \neq \emptyset.

Cites 1

Cites Definition 9

Proposition 2 03PJ

Let* G be a graph and let X, Y be two sets; then the relation X⊂YX \subset Y implies G⟨X⟩⊂G⟨Y⟩G\langle X \rangle \subset G\langle Y \rangle.

Cites 1 · Cited by 1

Cites II § 1. Collectivizing relations

Cited by Proposition 3, II § 4

This is an immediate consequence of the definitions and of C50 (§ 1, no. 4).

Corollary 03PK

If* A⊃pr1GA \supset \mathrm{pr}_1 G, we have G⟨A⟩=pr2GG\langle A \rangle = \mathrm{pr}_2 G.

Definition 4 03PL

Let* G be a graph and xx an object. The set G⟨{x}⟩G\langle \{x\} \rangle (which is sometimes denoted by G(x)G(x), by abuse of language) is called the section of G at xx.

Cites 1

Cites I § 5. Equalitarian theories

It follows immediately from C43 (Chapter I, § 5, no. 1) that the relation y∈G⟨{x}⟩y \in G\langle \{x\} \rangle is equivalent to (x,y)∈G(x, y) \in G. If G and G′G' are two graphs, the relation G⊂G′G \subset G' is thus equivalent to

(∀x)(G⟨{x}⟩⊂G′⟨{x}⟩).(\forall x)(G\langle \{x\} \rangle \subset G'\langle \{x\} \rangle).

If Γ=(G,A,B)\Gamma = (\mathrm{G}, \mathrm{A}, \mathrm{B}) is a correspondence between A and B, then for every x∈Ax \in \mathrm{A} the section of G at xx is also called the section of Γ\Gamma at xx and is denoted by Γ⟨{x}⟩\Gamma\langle\{x\}\rangle (or Γ(x)\Gamma(x)).

2. Inverse of a correspondence

Let G be a graph and A=pr1G\mathrm{A} = \mathrm{pr}_1\mathrm{G}, B=pr2G\mathrm{B} = \mathrm{pr}_2\mathrm{G} its projections. The relation (y,x)∈G(y, x) \in \mathrm{G} implies (x,y)∈B×A(x, y) \in \mathrm{B} \times \mathrm{A}; this relation therefore has a graph which consists of all ordered pairs (x,y)(x, y) such that (y,x)∈G(y, x) \in \mathrm{G}.

Definition 5 03PM

Let G be a graph. The graph whose elements are the ordered pairs (x,y)(x, y) such that (y,x)∈G(y, x) \in \mathrm{G} is called the inverse of G and is denoted by* G−1\overset{-1}{\mathrm{G}}.

For every set X, G−1⟨X⟩\overset{-1}{\mathrm{G}}\langle\mathrm{X}\rangle is called the inverse image of X under G.

¶ It is clear that the inverse of G−1\overset{-1}{\mathrm{G}} is G, and that

pr1G−1=pr2G,pr2G−1=pr1G.\mathrm{pr}_1\overset{-1}{\mathrm{G}} = \mathrm{pr}_2\mathrm{G}, \qquad \mathrm{pr}_2\overset{-1}{\mathrm{G}} = \mathrm{pr}_1\mathrm{G}.

In particular, if X, Y are two sets, we have

X×Y⏞−1=Y×X.\overset{-1}{\overbrace{\mathrm{X} \times \mathrm{Y}}} = \mathrm{Y} \times \mathrm{X}.

A graph G is said to be symmetric if G−1=G\overset{-1}{\mathrm{G}} = \mathrm{G}.

¶ Let Γ=(G,A,B)\Gamma = (\mathrm{G}, \mathrm{A}, \mathrm{B}) be a correspondence between A and B. Since pr1G−1⊂B\mathrm{pr}_1\overset{-1}{\mathrm{G}} \subset \mathrm{B} and pr2G−1⊂A\mathrm{pr}_2\overset{-1}{\mathrm{G}} \subset \mathrm{A}, the triple (G−1,B,A)(\overset{-1}{\mathrm{G}}, \mathrm{B}, \mathrm{A}) is a correspondence between B and A, called the inverse of the correspondence Γ\Gamma, and denoted by Γ−1\overset{-1}{\Gamma}. For every subset Y of B, the image Γ−1⟨Y⟩\overset{-1}{\Gamma}\langle\mathrm{Y}\rangle of Y under Γ−1\overset{-1}{\Gamma} is also called the inverse image of Y under Γ\Gamma. Clearly the inverse of Γ−1\overset{-1}{\Gamma} is Γ\Gamma.

3. Composition of two correspondences

Let G, G′ be two graphs. Let A denote the set pr1G\mathrm{pr}_1\mathrm{G} and let C denote the set pr2G′\mathrm{pr}_2\mathrm{G}'. The relation (∃y)((x,y)∈G(\exists y)((x, y) \in \mathrm{G} and (y,z)∈G′)(y, z) \in \mathrm{G}') implies that (x,z)∈A×C(x, z) \in \mathrm{A} \times \mathrm{C}, and therefore has a graph with respect to xx and zz.

Definition 6 03PN

Let G, G′ be two graphs. The graph (with respect to xx and zz) of the relation (∃y)((x,y)∈G(\exists y)((x, y) \in \mathrm{G} and (y,z)∈G′)(y, z) \in \mathrm{G}') is called the composition of G′ and G, and is denoted by* G′∘G\mathrm{G}' \circ \mathrm{G} (or sometimes by G′G\mathrm{G}'\mathrm{G}).

Cited by 1

Cited by Example, A I § 1

Proposition 3 03HK

Let* G, G′ be two graphs. The inverse of G′ ∘ G is then G−1∘G′−1\overset{-1}{G} \circ \overset{-1}{G'}.

Cited by 1

Cited by Definition 7

For the relation "(x,y)∈G(x, y) \in G and (y,z)∈G′(y, z) \in G'" is equivalent to

(z,y)∈G′−1 and (y,x)∈G−1.(z, y) \in \overset{-1}{G'} \text{ and } (y, x) \in \overset{-1}{G}.

Proposition 4 03HL

Let* G1G_1, G2G_2, G3G_3 *be graphs. Then

Cites 1

Cites I § 4. Quantified theories

(G3∘G2)∘G1=G3∘(G2∘G1).(G_3 \circ G_2) \circ G_1 = G_3 \circ (G_2 \circ G_1).

The relation (x,t)∈(G3∘G2)∘G1(x, t) \in (G_3 \circ G_2) \circ G_1 is equivalent to the relation

(∃y)((x,y)∈G1 and ∃z((y,z)∈G2 and (z,t)∈G3))(\exists y)((x, y) \in G_1 \text{ and } \exists z((y, z) \in G_2 \text{ and } (z, t) \in G_3))

and therefore (by C33 (Chapter I, § 4, no. 3)) to the relation

(1)

(∃y)(∃z)((x,y)∈G1 and (y,z)∈G2 and (z,t)∈G3).(\exists y)(\exists z)((x, y) \in G_1 \text{ and } (y, z) \in G_2 \text{ and } (z, t) \in G_3).

Similarly, the relation (x,t)∈G3∘(G2∘G1)(x, t) \in G_3 \circ (G_2 \circ G_1) is equivalent to

(2)

(∃z)(∃y)((x,y)∈G1 and (y,z)∈G2 and (z,t)∈G3).(\exists z)(\exists y)((x, y) \in G_1 \text{ and } (y, z) \in G_2 \text{ and } (z, t) \in G_3).

But the relations (1) and (2) are equivalent; hence the result.

¶ The graph G3∘(G2∘G1)G_3 \circ (G_2 \circ G_1) is denoted by G3∘G2∘G1G_3 \circ G_2 \circ G_1. Similarly, if G1G_1, G2G_2, G3G_3, G4G_4 are graphs, we put

G4∘(G3∘G2∘G1)=G4∘G3∘G2∘G1;G_4 \circ (G_3 \circ G_2 \circ G_1) = G_4 \circ G_3 \circ G_2 \circ G_1;

and so on.

Proposition 5 03HM

Let* G, G′ be graphs and let A *be a set. Then

Cites 1 · Cited by 1

Cites I § 4. Quantified theories

Cited by Definition 7

(G′∘G)⟨A⟩=G′⟨G⟨A⟩⟩.(G' \circ G)\langle A \rangle = G'\langle G\langle A \rangle\rangle.

For by virtue of C33 (Chapter I, § 4, no. 3) the relation z∈(G′∘G)⟨A⟩z \in (G' \circ G)\langle A \rangle is equivalent to

(∃y)((∃x)(x∈A and (x,y)∈G) and (y,z)∈G′)(\exists y)((\exists x)(x \in A \text{ and } (x, y) \in G) \text{ and } (y, z) \in G')

and is therefore equivalent to (∃y)(y∈G⟨A⟩(\exists y)(y \in G\langle A \rangle and (y,z)∈G′)(y, z) \in G'); hence the result.

¶ If G and G′ are two graphs, we have pr1(G′∘G)=G−1⟨pr1G′⟩\mathrm{pr}_1(G' \circ G) = \overset{-1}{G}\langle \mathrm{pr}_1 G'\rangle, and pr2(G′∘G)=G′⟨pr2G⟩\mathrm{pr}_2(G' \circ G) = G'\langle \mathrm{pr}_2 G\rangle. For example, to prove the second of these relations it is enough to note that the relation z∈pr2(G′∘G)z \in \mathrm{pr}_2(\mathrm{G}' \circ \mathrm{G}) is equivalent to (∃x)((x,z)∈G′∘G)(\exists x)((x, z) \in \mathrm{G}' \circ \mathrm{G}) and therefore to

(∃y)((∃x)((x,y)∈G)and(y,z)∈G′);(\exists y)((\exists x)((x, y) \in \mathrm{G}) \quad \text{and} \quad (y, z) \in \mathrm{G}');

but this is equivalent to z∈G′⟨pr2G⟩z \in \mathrm{G}'\langle \mathrm{pr}_2\mathrm{G}\rangle.

¶ If G is a graph and X a set such that X⊂pr1G\mathrm{X} \subset \mathrm{pr}_1\mathrm{G}, we have

X⊂G−1⟨G⟨X⟩⟩.\mathrm{X} \subset \overset{-1}{\mathrm{G}}\langle \mathrm{G}\langle \mathrm{X}\rangle\rangle.

For the relation x∈Xx \in \mathrm{X} implies by hypothesis that (∃y)((x,y)∈G)(\exists y)((x, y) \in \mathrm{G}); but (x,y)∈G(x, y) \in \mathrm{G} is equivalent to (y,x)∈G−1(y, x) \in \overset{-1}{\mathrm{G}}, and on the other hand (x,y)∈G(x, y) \in \mathrm{G} implies (∃z)(z∈X(\exists z)(z \in \mathrm{X} and (z,y)∈G)(z, y) \in \mathrm{G}); hence x∈Xx \in \mathrm{X} implies

(∃y)((∃z)(z∈Xand(z,y)∈G)and(y,x)∈G−1),(\exists y)((\exists z)(z \in \mathrm{X} \quad \text{and} \quad (z, y) \in \mathrm{G}) \quad \text{and} \quad (y, x) \in \overset{-1}{\mathrm{G}}),

that is to say, x∈G−1⟨G⟨X⟩⟩x \in \overset{-1}{\mathrm{G}}\langle \mathrm{G}\langle \mathrm{X}\rangle\rangle.

¶ It is clear that if G1\mathrm{G}_1, G2\mathrm{G}_2, G1′\mathrm{G}'_1, G2′\mathrm{G}'_2 are graphs, the relations G1⊂G2\mathrm{G}_1 \subset \mathrm{G}_2 and G1′⊂G2′\mathrm{G}'_1 \subset \mathrm{G}'_2 imply G1′∘G1⊂G2′∘G2\mathrm{G}'_1 \circ \mathrm{G}_1 \subset \mathrm{G}'_2 \circ \mathrm{G}_2.

¶ Let Γ=(G,A,B)\Gamma = (\mathrm{G}, \mathrm{A}, \mathrm{B}) and Γ′=(G′,B,C)\Gamma' = (\mathrm{G}', \mathrm{B}, \mathrm{C}) be two correspondences such that the target of Γ\Gamma is the same as the source of Γ′\Gamma'. From the above discussion we have pr1(G′∘G)⊂pr1G⊂A\mathrm{pr}_1(\mathrm{G}' \circ \mathrm{G}) \subset \mathrm{pr}_1\mathrm{G} \subset \mathrm{A} and pr2(G′∘G)⊂pr2G′⊂C\mathrm{pr}_2(\mathrm{G}' \circ \mathrm{G}) \subset \mathrm{pr}_2\mathrm{G}' \subset \mathrm{C}; hence we may state the following definition :

Definition 7 03PO

Let* Γ=(G,A,B)\Gamma = (\mathrm{G}, \mathrm{A}, \mathrm{B}) and Γ′=(G′,B,C)\Gamma' = (\mathrm{G}', \mathrm{B}, \mathrm{C}) be two correspondences such that the target of Γ\Gamma is the source of Γ′\Gamma'. Then the correspondence (G′∘G,A,C)(\mathrm{G}' \circ \mathrm{G}, \mathrm{A}, \mathrm{C}) is called the composition of Γ′\Gamma' and Γ\Gamma, and is denoted by Γ′∘Γ\Gamma' \circ \Gamma (or sometimes Γ′Γ\Gamma'\Gamma).

Cites 2

Cites Proposition 3, Proposition 5

It follows immediately from Proposition 5 that if X is a subset of A we have (Γ′∘Γ)⟨X⟩=Γ′⟨Γ⟨X⟩⟩(\Gamma' \circ \Gamma)\langle \mathrm{X}\rangle = \Gamma'\langle \Gamma\langle \mathrm{X}\rangle\rangle. Furthermore, since the target of Γ′−1\overset{-1}{\Gamma'} is the same as the source of Γ−1\overset{-1}{\Gamma}, the inverse of Γ′∘Γ\Gamma' \circ \Gamma is Γ−1∘Γ′−1\overset{-1}{\Gamma} \circ \overset{-1}{\Gamma'}, by Proposition 3.

Definition 8 03PP

If* A is a set, the set ΔA\Delta_\mathrm{A} of all objects of the form (x,x)(x, x), where x∈Ax \in \mathrm{A}, is called the diagonal of A×A\mathrm{A} \times \mathrm{A}.

Clearly we have pr1ΔA=pr2ΔA=A\mathrm{pr}_1\Delta_\mathrm{A} = \mathrm{pr}_2\Delta_\mathrm{A} = \mathrm{A}. The correspondence

IA=(ΔA,A,A)\mathrm{I}_\mathrm{A} = (\Delta_\mathrm{A}, \mathrm{A}, \mathrm{A})

is called the identity correspondence on A; it is its own inverse.

¶ If Γ\Gamma is a correspondence between A and B and if IA\mathrm{I}_\mathrm{A}, IB\mathrm{I}_\mathrm{B} are the identity correspondences on A, B, respectively, then Γ∘IA=IB∘Γ=Γ\Gamma \circ \mathrm{I}_\mathrm{A} = \mathrm{I}_\mathrm{B} \circ \Gamma = \Gamma.

4. Functions

Definition 9 03HN

A graph* F is said to be a functional graph if for each x there is at most one object which corresponds to x under F (Chapter I, § 5, no. 3). A correspondence f=(F,A,B)f = (\mathrm{F}, \mathrm{A}, \mathrm{B}) is said to be a function if its graph F is a functional graph and if its source A is equal to its domain pr1F\mathrm{pr}_1\mathrm{F}. In other words, a correspondence f=(F,A,B)f = (\mathrm{F}, \mathrm{A}, \mathrm{B}) is a function if for every x belonging to the source A of f the relation (x,y)∈F(x, y) \in \mathrm{F} is functional in y (Chapter I, § 5, no. 3); the unique object which corresponds to x under f is called the value of f at the element x of A, and is denoted by f(x)f(x) (or fxf_x, or F(x)\mathrm{F}(x), or Fx\mathrm{F}_x).

Cites 1 · Cited by 1

Cites I § 5. Equalitarian theories

Cited by Remark 2

If ff is a function, F its graph, and xx an element of the domain of ff, the relation y=f(x)y = f(x) is then equivalent to (x,y)∈F(x, y) \in \mathrm{F} (Chapter I, § 5, no. 3, criterion C46).

Remark 03HO

The reader should beware of the confusion which may arise from the simultaneous use of the notations f(x)f(x) and f(X)f(\mathrm{X}) (synonymous with f⟨X⟩f\langle \mathrm{X} \rangle) introduced in Definition 3 (cf. Exercise 11).

Let A and B be two sets; a mapping of A into B is a function ff whose source (which is equal to its domain) is equal to A and whose target is equal to B; such a function is also said to be defined on A and to take its values in B.

Instead of the phrase "let ff be a mapping of A into B", the following phrases are often used : "let f:A→Bf : \mathrm{A} \to \mathrm{B} be a mapping" or even "let f:A→Bf : \mathrm{A} \to \mathrm{B}". To simplify the presentation of an argument involving several mappings, we use diagrams such as

fgCiA→B↗↘E↘D↗hj\begin{array}{ccccccccc} & f & & & g & \mathrm{C} & & i & \\ \mathrm{A} & \to & \mathrm{B} & \nearrow & & & \searrow & & \mathrm{E} \\ & & & \searrow & & \mathrm{D} & \nearrow & & \\ & & & h & & & j & & \end{array}

in which a group of signs such as A→fB\mathrm{A} \overset{f}{\to} \mathrm{B} is to be interpreted as meaning that ff is a mapping of A into B.

A function ff defined on A is said to transform x into f(x)f(x) (for all x∈Ax \in \mathrm{A}); f(x)f(x) is called the transform of x by f or (by abuse of language) the image of xx under ff.

¶ Under certain circumstances, a functional graph is called a family; the domain is then called the index set, and the range is called (by abuse of language) the set of elements of the family. It is mainly in this connection that the indicial notation fxf_x is used to denote the value of ff at the element xx. When the index set is the product of two sets, we often speak of a double family.

¶ Likewise a function whose target is E is sometimes called a family of elements of E. If every element of E is a subset of a set F, we speak of a family of subsets of F.

Throughout this series we shall often use the word “function” in place of “functional graph”.

Examples of functions

Cites 2

Cites Definition 3, Exercise 11, § 3

Example 1 03V1

The empty set is a functional graph. Every function whose graph is empty has domain and range equal to the empty set. Among such functions, the one whose target is empty (i.e., the function (∅,∅,∅)(\varnothing,\varnothing,\varnothing)) is called the empty function.

Cited by 1

Cited by Definition 1, II § 5

Example 2 03V2

Let A be a set. The identity correspondence of A (no. 3) is a function, called the identity mapping of A.

Thus with every set A there is associated a family, defined by the identity mapping of A, whose index set is A and whose set of elements is A. By abuse of language, a set is sometimes referred to as a “family”, in which case it is the family thus associated with the set in question.

A function ff is said to be constant if for all xx and x′x' in the domain of ff we have f(x)=f(x′)f(x)=f(x').

¶ Let ff be a mapping of a set E into E. An element xx of E is said to be fixed under ff if f(x)=xf(x)=x.

5. Restrictions and extensions of functions

Two functions ff and gg are said to agree (or coincide) on a set E if E is contained in the domains of ff and gg and if f(x)=g(x)f(x)=g(x) for all x∈Ex\in E. Two functions which have the same graph agree on their domain. To say that f=gf=g is to say that ff and gg have the same domain A and the same target B, and that they agree on A.

¶ Let f=(F,A,B)f=(F,A,B) and g=(G,C,D)g=(G,C,D) be two functions. To say that F⊂GF\subset G is to say that the domain A of ff is contained in the domain C of gg and that gg agrees with ff on A. If also B⊂DB\subset D, then gg is said to be an extension of ff (more precisely, an extension of ff to C), and gg is said to extend ff (to C). When gg is called a family of elements of D, ff is said to be a subfamily of gg.

¶ Let ff be a function and let A be a subset of the domain of ff. It is immediate that the relation “x∈Ax\in A and y=f(x)y=f(x)” has a graph G with respect to xx and yy, that this graph is functional, and that A is its domain; the function whose graph is G, which has the same target as ff, is called the restriction of ff to AA, and is sometimes denoted by f∣Af|A. A function is an extension of any of its restrictions. If two functions ff, gg have the same target and agree on a set EE, then their restrictions to EE are equal.

6. Definition of a function by means of a term

C54. Let TT, AA be two terms and let xx, yy be distinct letters. Suppose that xx does not appear in AA and that yy does not appear in either TT or AA. Let RR be the relation “x∈Ax \in A and y=Ty = T”. The relation RR has a graph FF with respect to the letters xx and yy. This graph is functional; its first projection is AA, and its second projection is the set of objects of the form TT for x∈Ax \in A (§ 1, no. 6). For every x∈Ax \in A we have F(x)=TF(x) = T.

Let BB be the set of objects of the form TT for x∈Ax \in A. Then

R⟹((x,y)∈A×B); R \Longrightarrow ((x,y) \in A \times B);

since the assembly denoted by A×BA \times B contains neither xx nor yy, RR has a graph FF with respect to the letters xx and yy (no. 1). It is clear that the relation

(x,y)∈Fand(x,y′)∈F (x,y) \in F \quad\text{and}\quad (x,y') \in F

implies y=y′y = y', and hence FF is a functional graph. The remaining statements are evident.

¶ If CC is a set which contains the set BB of objects of the form TT for x∈Ax \in A (where yy does not appear in CC), then the function (F,A,C)(F,A,C) is also denoted by the notation x→Tx \to T (x∈Ax \in A, T∈CT \in C); the corresponding assembly in formal mathematics contains neither xx nor yy and does not depend on the choice of the letter yy satisfying the above conditions. When the context is sufficiently explicit we shall permit ourselves the notations x→T(x∈A)x \to T(x \in A), (T)x∈A(T)_{x \in A}, or x→Tx \to T, and sometimes simply TT or (T)(T). Thus we may speak of “the function x3x^3”, if the context indicates clearly that we mean the mapping x→x3x \to x^3 of the set of complex numbers into itself. \

Examples

Example 1 03S3

If ff is a mapping of AA into BB, the function ff is equal to the function x→f(x)x \to f(x) (x∈Ax \in A, f(x)∈Bf(x) \in B), which is written simply as x→f(x)x \to f(x) or also (fx)x∈A(f_x)_{x \in A} (the latter notation is especially associated with the phrase “family of elements” instead of “function”).

Example 2 03S4

Let G be a set of ordered pairs. The functions

z→pr1z(z∈G, pr1z∈pr1G)z \to \mathrm{pr}_1 z \quad (z \in \mathrm{G}, \ \mathrm{pr}_1 z \in \mathrm{pr}_1 \mathrm{G})

and

z→pr2z(z∈G, pr2z∈pr2G)z \to \mathrm{pr}_2 z \quad (z \in \mathrm{G}, \ \mathrm{pr}_2 z \in \mathrm{pr}_2 \mathrm{G})

are called respectively the first and second coordinate functions on G; they are denoted by pr1\mathrm{pr}_1 and pr2\mathrm{pr}_2 when there is no risk of confusion.

7. Composition of two functions. Inverse function

Proposition 6 03HP

If f is a mapping of* A into B, and g is a mapping of B into C, then g∘fg \circ f is a mapping of A into C.

Let F, G be the graphs of ff, gg, respectively, and let us show that G∘F\mathrm{G} \circ \mathrm{F} is a functional graph. Let xx, zz, z′z' be objects such that (x,z)∈G∘F(x, z) \in \mathrm{G} \circ \mathrm{F}, (x,z′)∈G∘F(x, z') \in \mathrm{G} \circ \mathrm{F}. There exist objects yy, y′y' such that

(x,y)∈F,(x,y′)∈F,(y,z)∈G,(y′,z′)∈G.(x, y) \in \mathrm{F}, \qquad (x, y') \in \mathrm{F}, \qquad (y, z) \in \mathrm{G}, \qquad (y', z') \in \mathrm{G}.

Since F is a functional graph, we have y=y′y = y' and hence (y,z′)∈G(y, z') \in \mathrm{G}. Since G is a functional graph, it follows that z=z′z = z', which proves our assertion. Furthermore, the domain of g∘fg \circ f is evidently A, and the proof is complete.

¶ The function g∘fg \circ f may also be written as x→g(f(x))x \to g(f(x)), or as gfgf when there is no risk of confusion.

Definition 10 03HQ

Let f be a mapping of* A into B. The mapping f is said to be injective, or an injection, if any two distinct elements of A have distinct images under f. The mapping f is said to be surjective, or a surjection, if f(A)=Bf(\mathrm{A}) = \mathrm{B}. *If f is both injective and surjective, it is said to be bijective, or a bijection.

Instead of saying that ff is surjective, we sometimes say that ff is a mapping of A onto B, or that it is a parametric representation of B by means of A (in which case A is called the set of parameters of the representation, and the elements of A are called parameters). If ff is bijective, we sometimes say that f puts A and B in one-to-one correspondence. A bijection of A onto A is called a permutation of A.

Examples

Example 1 03S5

If A⊂B\mathrm{A} \subset \mathrm{B}, the mapping of A into B whose graph is the diagonal of A is injective and is called the canonical mapping or the canonical injection (or simply the injection) of A into B.

Example 2 03S6

Let A be a set. The mapping x→(x,x)x \to (x, x) of A into the diagonal ΔA\Delta_\mathbf{A} of A×A\mathrm{A} \times \mathrm{A} is a bijective mapping, called the diagonal mapping of A.

Example 3 03S7

Let G be a set of ordered pairs. The mapping pr1\mathrm{pr}_1 (resp. pr2\mathrm{pr}_2) of G into pr1G\mathrm{pr}_1\mathrm{G} (resp. pr2G\mathrm{pr}_2\mathrm{G}) is surjective; pr1\mathrm{pr}_1 is injective if and only if G is a functional graph.

Example 4 03S8

Let G be a set of ordered pairs. The mapping

z→(pr2z, pr1z)z \to (\mathrm{pr}_2 z,\ \mathrm{pr}_1 z)

of G into G−1\overset{-1}{\mathrm{G}} is a bijection (called the canonical bijection).

Example 5 03S9

Let A be a set and bb an object. The mapping x→(x, b)x \to (x,\ b) of A into A×{b}\mathrm{A} \times \{b\} is a bijection.

Proposition 7 03PQ

Let ff be a mapping of* A into B. Then f−1\overset{-1}{f} *is a function if and only if ff is bijective.

If f−1\overset{-1}{f} is a function, its source B is equal to its domain, i.e., to f(A)f(\mathrm{A}); hence ff is surjective. To show that ff is injective, let xx and yy be two elements of A such that f(x)=f(y)f(x) = f(y). If F denotes the graph of ff, we have

(f(x), x)∈F−1and(f(y), y)∈F−1,(f(x),\ x) \in \overset{-1}{\mathrm{F}} \qquad \text{and} \qquad (f(y),\ y) \in \overset{-1}{\mathrm{F}},

hence

(f(x), y)∈F−1,(f(x),\ y) \in \overset{-1}{\mathrm{F}},

so that x=yx = y, which proves the assertion.

¶ Conversely, if ff is bijective, it is immediate that F−1\overset{-1}{\mathrm{F}} is functional and that the domain of f−1\overset{-1}{f} is equal to B.

When ff is bijective, f−1\overset{-1}{f} is called the inverse mapping of ff; f−1\overset{-1}{f} is bijective, f−1∘f\overset{-1}{f} \circ f is the identity mapping of A, and f∘f−1f \circ \overset{-1}{f} is the identity mapping of B.

¶ If a permutation is the same as its inverse, it is said to be involutory.

Remark 03HR

Let ff be a mapping of A into B. For each subset X of A we have (no. 3) X⊂f−1⟨f⟨X⟩⟩\mathrm{X} \subset \overset{-1}{f}\langle f\langle \mathrm{X}\rangle\rangle. Furthermore, for each subset Y of B we have f⟨f−1⟨Y⟩⟩⊂Yf\langle \overset{-1}{f}\langle \mathrm{Y}\rangle\rangle \subset \mathrm{Y}, for the relation y∈f⟨f−1⟨Y⟩⟩y \in f\langle \overset{-1}{f}\langle \mathrm{Y}\rangle\rangle is equivalent to

(∃x)((∃z)(z∈Y and z=f(x)) and y=f(x))(\exists x)((\exists z)(z \in \mathrm{Y} \text{ and } z = f(x)) \text{ and } y = f(x))

and therefore implies the relation (∃z)(z∈Y and y=z)(\exists z)(z \in \mathrm{Y} \text{ and } y = z) and consequently also the relation y∈Yy \in \mathrm{Y}.

If ff is surjective, we have f⟨f−1⟨Y⟩⟩=Yf\langle \overset{-1}{f}\langle \mathrm{Y}\rangle\rangle = \mathrm{Y} for every subset Y of B, for the relation y∈Y⊂By \in \mathrm{Y} \subset \mathrm{B} implies by hypothesis the relation (∃x)(y=f(x))(\exists x)(y = f(x)) and therefore also (∃x)(y∈Y and y=f(x))(\exists x)(y \in \mathrm{Y} \text{ and } y = f(x)); but "y∈Yy \in \mathrm{Y} and y=f(x)y = f(x)" implies (∃z)(z∈Y and z=f(x))(\exists z)(z \in \mathrm{Y} \text{ and } z = f(x)), and our assertion follows.

If ff is injective, we have f−1⟨f⟨X⟩⟩=X\overset{-1}{f}\langle f\langle X\rangle\rangle = X for every subset X of A. For the relation x∈f−1⟨f⟨X⟩⟩x \in \overset{-1}{f}\langle f\langle X\rangle\rangle is equivalent to f(x)∈f⟨X⟩f(x) \in f\langle X\rangle, hence to

(∃z)(z∈X and f(z)=f(x));(\exists z)(z \in X \text{ and } f(z) = f(x));

but the hypothesis means that f(z)=f(x)f(z) = f(x) implies z=xz = x, hence x∈f−1⟨f⟨X⟩⟩x \in \overset{-1}{f}\langle f\langle X\rangle\rangle implies x∈Xx \in X.

8. Retractions and sections

Proposition 8 03HS

Let ff be a mapping of* A into B. If there exists a mapping rr (resp. ss) of B into A such that r∘fr \circ f (resp. f∘sf \circ s) is the identity mapping of A (resp. B), then ff is injective (resp. surjective). Conversely, if ff is surjective, there exists a mapping ss of B into A such that f∘sf \circ s is the identity mapping of B. If ff is injective and if A ≠∅\neq \emptyset, there exists a mapping rr of B into A such that r∘fr \circ f is the identity mapping of A.

Cited by 5

Cited by Proposition 2, A I § 7, Theorem 1, Corollary, II § 5, Proposition 12, III § 3, Proposition 3, III § 3

If there exists a mapping rr of B into A such that r∘fr \circ f is the identity mapping of A, then the equality f(x)=f(y)f(x) = f(y), where x∈Ax \in A and y∈Ay \in A, implies x=r(f(x))=r(f(y))=yx = r(f(x)) = r(f(y)) = y, and so ff is injective. If there exists a mapping ss of B into A such that f∘sf \circ s is the identity mapping of B, we have B=f(s(B))⊂f(A)⊂BB = f(s(B)) \subset f(A) \subset B, so that ff is surjective. If ff is surjective, let T denote the term τy(y∈A and f(y)=x)\tau_y(y \in A \text{ and } f(y) = x). We have f(T)=xf(T) = x for x∈Bx \in B; if ss denotes the mapping x→Tx \to T (x∈B,T∈A)(x \in B, T \in A), then f∘sf \circ s is the identity mapping of B. Finally, suppose that ff is injective and that A ≠∅\neq \emptyset, and let aa be an element of A. The relation

“(y∈A and x=f(y)) or (y=a and x∈B−f(A))”\text{“}(y \in A \text{ and } x = f(y)) \text{ or } (y = a \text{ and } x \in B - f(A))\text{''}

implies (x,y)∈B×A(x, y) \in B \times A and therefore has a graph R with respect to the letters xx, yy. This graph is functional by reason of the hypothesis on ff, and has B as its domain; and we have R(x)=aR(x) = a if x∈B−f(A)x \in B - f(A), and f(R(x))=xf(R(x)) = x if x∈f(A)x \in f(A). Hence the function r=(R,B,A)r = (R, B, A) is such that r∘fr \circ f is the identity mapping of A.

Corollary 03HT

Let* A, B be sets, let ff be a mapping of A into B, and let gg be a mapping of B into A. If g∘fg \circ f is the identity mapping of A and if f∘gf \circ g is the identity mapping of B, *then ff and gg are bijective, and g=f−1g = \overset{-1}{f}.

Cited by 1

Cited by Remark, IV § 2

Definition 11 03HU

Let ff be an injective* (resp. surjective) mapping of A into B. Any mapping rr (resp. ss) of B into A such that r∘fr \circ f (resp. f∘sf \circ s) is the identity mapping of A (resp. B) is called a retraction (resp. section) *of ff.

Cited by 6

Cited by Example, A I § 2, Proposition 15, A II § 1, Corollary, II § 5, Proposition 2, II § 5, II § 6. Equivalence relations, Proposition 9, TG I § 3

Instead of retraction (resp. section) the phrase left-inverse (resp. right-inverse) is sometimes used.

¶ If ff is injective (resp. surjective) and if rr (resp. ss) is a retraction (resp. section) of ff, then ff is a section (resp. retraction) of rr (resp. ss). Hence a retraction is surjective and a section is injective.

¶ If ff is surjective and if ss, s′s' are two sections of ff such that s(B)=s′(B)s(\mathrm{B}) = s'(\mathrm{B}), then s=s′s = s'; for if x∈Bx \in \mathrm{B}, there exists y∈By \in \mathrm{B} such that s(x)=s′(y)s(x) = s'(y), and we have x=f(s(x))=f(s′(y))=yx = f(s(x)) = f(s'(y)) = y, so that s(x)=s′(x)s(x) = s'(x) and consequently s=s′s = s'. Thus a section ss is uniquely determined by the set s(B)s(\mathrm{B}). By abuse of language, the set s(B)s(\mathrm{B}) is sometimes called a section of ff.

Theorem 1 03PR

Let ff be a mapping of* A into B, let f′f' be a mapping of B into C, *and let f′′=f′∘ff'' = f' \circ f. Then :

Cites 1 · Cited by 2

Cites Proposition 8

Cited by III § 3. Equipotent sets. Cardinals, Definition 1, III § 3

(a) If ff and f′f' are injections, then f′′f'' is an injection. If rr, r′r' are retractions of ff, f′f', respectively, then r∘r′r \circ r' is a retraction of f′′f''.

(b) If ff and f′f' are surjections, then f′′f'' is a surjection. If ss, s′s' are sections of ff, f′f', respectively, then s∘s′s \circ s' is a section of f′′f''.

(c) If f′′f'' is an injection, then ff is an injection. If r′′r'' is a retraction of f′′f'', then r′′∘f′r'' \circ f' is a retraction of ff.

(d) If f′′f'' is a surjection, then f′f' is a surjection. If s′′s'' is a section of f′′f'', then f∘s′′f \circ s'' is a section of f′f'.

(e) If f′′f'' is a surjection and f′f' an injection, then ff is a surjection. If s′′s'' is a section of f′′f'', then s′′∘f′s'' \circ f' is a section of ff.

(f) If f′′f'' is an injection and ff a surjection, then f′f' is an injection. If r′′r'' is a retraction of f′′f'', then f∘r′′f \circ r'' is a retraction of f′f'.

For every set E let IE\mathrm{I_E} denote the identity mapping of E.

(a) We have r∘f=IAr \circ f = \mathrm{I_A} and r′∘f′=IBr' \circ f' = \mathrm{I_B}, hence

(r∘r′)∘(f′∘f)=r∘IB∘f=r∘f=IA.(r \circ r') \circ (f' \circ f) = r \circ \mathrm{I_B} \circ f = r \circ f = \mathrm{I_A}.

If ff and f′f' are injections, then f′′f'' is an injection, by Proposition 8 if A≠∅\mathrm{A} \neq \emptyset, and trivially if A=∅\mathrm{A} = \emptyset.

(b) We have f∘s=IBf \circ s = \mathrm{I_B} and f′∘s′=ICf' \circ s' = \mathrm{I_C}, hence

(f′∘f)(s∘s′)=f′∘IB∘s′=f′∘s′=IC.(f' \circ f)(s \circ s') = f' \circ \mathrm{I_B} \circ s' = f' \circ s' = \mathrm{I_C}.

If ff and f′f' are surjections, f′′f'' is then a surjection by Proposition 8.

(c) We have r′′∘f′′=IAr'' \circ f'' = \mathrm{I_A}, hence (r′′∘f′)∘f=r′′∘f′′=IA(r'' \circ f') \circ f = r'' \circ f'' = \mathrm{I_A}. If f′′f'' is an injection, then ff is an injection, by Proposition 8 if A≠∅\mathrm{A} \neq \emptyset, and trivially if A=∅\mathrm{A} = \emptyset.

(d) We have f′′∘s′′=ICf'' \circ s'' = \mathrm{I_C}, hence f′∘(f∘s′′)=f′′∘s′′=ICf' \circ (f \circ s'') = f'' \circ s'' = \mathrm{I_C}. If f′′f'' is a surjection, then f′f' is a surjection by Proposition 8.

(e) We have f′′∘s′′=ICf'' \circ s'' = \mathrm{I_C}, and f′f' is a bijection by (d). Hence

f∘(s′′∘f′)  =  (f−1′∘f′)∘f∘(s′′∘f′)  =  f−1′∘(f′′∘s′′)∘f′f \circ (s'' \circ f') \;=\; (\overset{-1}{f}{}' \circ f') \circ f \circ (s'' \circ f') \;=\; \overset{-1}{f}{}' \circ (f'' \circ s'') \circ f'
=f−1′∘IC∘f′=f−1′∘f′=IB.= \overset{-1}{f}{}' \circ \mathrm{I_C} \circ f' = \overset{-1}{f}{}' \circ f' = \mathrm{I_B}.

If f′′f'' is a surjection and f′f' an injection, ff is then a surjection by Proposition 8.

(f) We have r′′∘f′′=IAr'' \circ f'' = \mathrm{I_A}, and ff is a bijection by (c). Hence

(f∘r′′)∘f′=(f∘r′′)∘f′∘(f∘f−1)=f∘(r′′∘f′′)∘f−1=f∘IA∘f−1(f \circ r'') \circ f' = (f \circ r'') \circ f' \circ (f \circ \overset{-1}{f}) = f \circ (r'' \circ f'') \circ \overset{-1}{f} = f \circ \mathrm{I_A} \circ \overset{-1}{f}
=f∘f−1=IB.= f \circ \overset{-1}{f} = \mathrm{I_B}.

If f′′f'' is an injection and ff a surjection, then f′f' is an injection, by Proposition 8 if A≠∅\mathrm{A} \neq \emptyset, and trivially if A=∅\mathrm{A} = \emptyset (for then we have B=f⟨A⟩=∅\mathrm{B} = f\langle \mathrm{A}\rangle = \emptyset).

Proposition 9 03PS

(a) Let E, F, G be sets, let gg be a mapping of E onto F and ff a mapping of E into G. Then there exists a mapping hh of F into G such that f=h∘gf = h \circ g if and only if the relation g(x)=g(y)g(x) = g(y) (where x∈Ex \in \mathrm{E}, y∈Ey \in \mathrm{E}*) implies the relation* f(x)=f(y)f(x) = f(y). The mapping hh is then uniquely determined by ff; if ss is a section of gg, we have h=f∘sh = f \circ s.

E↘fg↓↑sF⟶hG↗fEr↓↑gG⟶hF\begin{array}{ccc} \mathrm{E} & \overset{f}{\searrow} & \\ {\scriptstyle g}\big\downarrow\big\uparrow{\scriptstyle s} & & \\ \mathrm{F} & \underset{h}{\longrightarrow} & \mathrm{G} \end{array} \qquad\qquad \begin{array}{ccc} & \overset{f}{\nearrow} & \mathrm{E} \\ & & {\scriptstyle r}\big\downarrow\big\uparrow{\scriptstyle g} \\ \mathrm{G} & \underset{h}{\longrightarrow} & \mathrm{F} \end{array}
(a)(b)\text{(a)} \qquad\qquad\qquad \text{(b)}
Cited by 1

Cited by II § 6. Equivalence relations

(b) Let E, F, G be sets, let gg be an injection of F into E, and let ff be a mapping of G into E. Then there exists a mapping hh of G into F such that f=g∘hf = g \circ h if and only if f(G)⊂f(F)f(\mathrm{G}) \subset f(\mathrm{F}). The mapping hh is uniquely determined by ff; if rr is a retraction of gg, we have h=r∘fh = r \circ f.

(a) If f=h∘gf = h \circ g, the relation g(x)=g(y)g(x) = g(y) (where x∈Ex \in \mathrm{E}, y∈Ey \in \mathrm{E}) clearly implies f(x)=f(y)f(x) = f(y). And for every section ss of gg we have

h=h∘(g∘s)=f∘s,h = h \circ (g \circ s) = f \circ s,

which shows that hh is uniquely determined by ff. Conversely, suppose that the relation g(x)=g(y)g(x) = g(y) implies f(x)=f(y)f(x) = f(y); let ss be a section of gg, and let h=f∘sh = f \circ s; then for every x∈Ex \in E we have g(s(g(x)))=g(x)g(s(g(x))) = g(x), hence f(s(g(x)))=f(x)f(s(g(x))) = f(x), that is, h(g(x))=f(x)h(g(x)) = f(x) and therefore f=h∘gf = h \circ g.

(b) If f=g∘hf = g \circ h, then clearly f(G)⊂f(F)f(G) \subset f(F), and for every retraction rr of gg we have h=(r∘g)∘h=r∘fh = (r \circ g) \circ h = r \circ f, which shows that hh is uniquely determined by ff. Conversely, suppose that f(G)⊂f(F)f(G) \subset f(F); let rr be a retraction of gg, and put h=r∘fh = r \circ f; for every x∈Gx \in G, there exists y∈Fy \in F such that f(x)=g(y)f(x) = g(y), so that

g(h(x))=g(r(f(x)))=g(r(g(y)))=g(y)=f(x)g(h(x)) = g(r(f(x))) = g(r(g(y))) = g(y) = f(x)

and therefore f=g∘hf = g \circ h.

9. Functions of two arguments

A function of two arguments is a function whose domain is a set of ordered pairs (or, equivalently, a subset of a product). Let ff be such a function; if (x,y)(x, y) is an element of the domain of ff, the value f((x,y))f((x, y)) of ff at the element (x,y)(x, y) is generally denoted by f(x,y)f(x, y).

¶ Let ff be a function of two arguments, D its domain, and C its target. For each yy let AyA_y be the set of all xx such that (x,y)∈D(x, y) \in D (that is, the section of D−1\overset{-1}{D} at yy (no. 1)). The mapping

x→f(x,y)(x∈Ay,f(x,y)∈C)x \to f(x, y) \quad (x \in A_y, f(x, y) \in C)

is called the partial mapping defined by ff, with respect to the value yy of the second argument, and is denoted by f(∙,y)f(\bullet, y), or f( ,y)f(\ , y) (or sometimes fyf_y); we have f(∙,y)(x)=f(x,y)f(\bullet, y)(x) = f(x, y) for all (x,y)∈D(x, y) \in D. Similarly, for each xx let BxB_x be the set of all yy such that (x,y)∈D(x, y) \in D. The mapping

y→f(x,y)(y∈Bx,f(x,y)∈C)y \to f(x, y) \quad (y \in B_x, f(x, y) \in C)

is called the partial mapping defined by ff, with respect to the value xx of the first argument, and is denoted by f(x,∙)f(x, \bullet), or f(x, )f(x, \ ) (or sometimes fxf_x); we have f(x,∙)(y)=f(x,y)f(x, \bullet)(y) = f(x, y) for all (x,y)∈D(x, y) \in D.

¶ If for every yy (resp. xx) the partial mapping f(∙,y)f(\bullet, y) (resp. f(x,∙)f(x, \bullet)) is a constant mapping, we say that ff does not depend on its first (resp. second) argument; this means therefore that f(x,y)=f(x′,y)f(x, y) = f(x', y) whenever (x,y)(x, y) and (x′,y)(x', y) belong to D (resp. f(x,y)=f(x,y′)f(x, y) = f(x, y') whenever (x,y)(x, y) and (x,y′)(x, y') belong to D). For each yy belonging to the second projection of D let g(y)g(y) denote the common value of the f(x,y)f(x, y) for x∈Ayx \in A_y; the mapping y→g(y)y \to g(y) is then a mapping of pr2D\mathrm{pr}_2 D into C such that

g(y)=f(x,y)for all (x,y)∈D.g(y) = f(x, y) \qquad \text{for all } (x, y) \in D.

¶ Conversely, let gg be a mapping of a set B into a set C, and let A be any set. Then the mapping (x,y)→g(y)(x, y) \to g(y) of A×BA \times B into C does not depend on its first argument.

¶ Let uu be a mapping of A into C and vv a mapping of B into D. The mapping z→(u(pr1z),v(pr2z))z \to (u(\mathrm{pr}_1 z), v(\mathrm{pr}_2 z)) of A×BA \times B into C×DC \times D is called the (canonical) extension of uu and vv to the product sets, or simply the product of uu and vv (if there is no risk of confusion); its range is u⟨A⟩×v⟨B⟩u\langle A\rangle \times v\langle B\rangle. It is denoted by u×vu \times v. If uu and vv are injective (resp. surjective) then u×vu \times v is injective (resp. surjective). If uu and vv are bijective, then u×vu \times v is bijective, and its inverse mapping is u−1×v−1\overset{-1}{u} \times \overset{-1}{v}. If u′u' is a mapping of C into a set E and if v′v' is a mapping of D into a set F, we have

(u′×v′)∘(u×v)=(u′∘u)×(v′∘v).(u' \times v') \circ (u \times v) = (u' \circ u) \times (v' \circ v).

If U, V are the graphs of uu, vv respectively, the graph W of u×vu \times v is the set of ordered pairs ((x,y),(z,t))((x,y), (z, t)) of (A×B)×(C×D)(A \times B) \times (C \times D) such that (x,z)∈U(x, z) \in U and (y,t)∈V(y, t) \in V; the mapping ((x,y),(z,t))→((x,z),(y,t))((x, y), (z, t)) \to ((x, z), (y, t)) puts W in one-to-one correspondence with the product U×VU \times V (a subset of (A×C)×(B×D)(A \times C) \times (B \times D)) (cf. §5, no. 5).

Exercises

See the exercises for § 3.

Transcribed from the printed edition, 2004, Springer. Not translated.