to be exact, it is necessary and sufficient that, for every A-module F, the sequence

(where $\bar{u} = \mathrm{Hom}(u, 1_F)$, $\bar{v} = \mathrm{Hom}(v, 1_F)$) be exact.

Suppose that sequence (1) is exact. If $w \in \mathrm{Hom}(E'', F)$ and $\bar{v}(w) = w \circ v = 0$, then $w = 0$ since $v$ is surjective. Sequence (2) is therefore exact at $\mathrm{Hom}(E'', F)$. We show that it is exact at $\mathrm{Hom}(E, F)$. $\bar{u} \circ \bar{v} = \mathrm{Hom}(v \circ u, 1_F)$ (§ 1, no. 2, formula (10)) and $v \circ u = 0$ since sequence (1) is exact at E. Therefore $\bar{u} \circ \bar{v} = 0$, that is $\mathrm{Im}(\bar{v}) \subset \mathrm{Ker}(\bar{u})$. On the other hand, if $w \in \mathrm{Ker}(\bar{u})$, then $w \circ u = 0$ and hence $\mathrm{Ker}(w) \supset \mathrm{Im}(u)$. But as sequence (1) is exact at E, $\mathrm{Im}(u) = \mathrm{Ker}(v)$ and hence $\mathrm{Ker}(w) \supset \mathrm{Ker}(v)$; as $v$ is surjective, it follows from § 1, no. 3, Remark that there exists a $w' \in \mathrm{Hom}(E'', F)$ such that w = w' \circ v = \bar{v}(w'). Therefore Ker(\bar{u}) \subset \operatorname{Im}(\bar{v}), which completes the proof that sequence (2) is exact.

Conversely, suppose that (2) is exact for every A-module F. As $\bar{u} \circ \bar{v} = \operatorname{Hom}(v \circ u, 1_F) = 0, w \circ v \circ u = 0$ for every homomorphism $w : E'' \to F$. Taking $F = E''$ and $w = 1_{E''}$, it is seen first that $v \circ u = 0$ and hence $u(E') \subset \operatorname{Ker}(v)$. Now take $F = \operatorname{Coker}(u)$ and let $\phi : E \to F = E/u(E')$ be the canonical mapping. Then $\bar{u}(\phi) = \phi \circ u = 0$ by definition and hence there exists a $\psi \in \operatorname{Hom}(E'', F)$ such that $\phi = \bar{v}(\psi) = \psi \circ v$; this obviously implies $u(E') = \operatorname{Ker}(\phi) \supset \operatorname{Ker}(v)$, which proves that sequence (1) is exact at E. Finally, let $\theta$ be the canonical homomorphism of $E''$ onto $F = E''/v(E)$; then $\bar{v}(\theta) = \theta \circ v = 0$, hence $\theta = 0$; therefore, $F = \{0\}$ and $v$ is surjective. Sequence (1) is therefore exact at $E''$.

It suffices to apply Theorem 1 to the case where $E'' = \{0\}$ (resp. $E' = \{0\}$ resp. $E'' = E$ and $v = 1_E$).

Note that starting from an exact sequence

the corresponding sequence

is not necessarily exact, in other words, the homomorphism $\bar{u}$ is not necessarily surjective. If $E'$ is identified with a submodule of E, this means that a linear mapping of $E'$ into F cannot always be extended to a linear mapping of E into F (Exercises 11 and 12). However:

(3)

splits (in other words, if $u(E')$ is a direct factor of E) the sequence

(4)

is exact and splits. Conversely, if, for every A-module F, sequence (4) is exact, sequence (3) splits.

If the exact sequence (3) splits, there exists a linear retraction $u' : E \to E'$ associated with $u$ (§ 1, no. 9, Proposition 15); if

the fact that $u' \circ u$ is the identity implies that $\bar{u} \circ \bar{u}'$ is the identity ($§ 1$, no. 2, formula (10)) and hence the first assertion follows from $§ 1$, no. 9, Proposition 15. Conversely, suppose sequence (4) is exact for $F = E'$. Then there exists an element $f \in \mathrm{Hom}(E, E')$ such that $f \circ u = 1_{E'}$, and the conclusion follows from $§ 1$, no. 9, Proposition 15.

Note that the first assertion of Proposition 1 can also be considered as a special case of $§ 1$, no. 6, Corollary 1 to Proposition 6, canonically identifying $\mathrm{Hom}(E', F) \oplus \mathrm{Hom}(E'', F)$ with $\mathrm{Hom}(E' \oplus E'', F)$ by means of the $\mathbf{Z}$-linear mapping $\mathrm{Hom}(p', 1_F) + \mathrm{Hom}(p'', 1_F)$, where $p': E' \oplus E'' \to E'$ and $p'': E' \oplus E'' \to E''$ are the canonical projections.

to be exact, it is necessary and sufficient that, for every $A$-module $E$, the sequence

(where $\bar{u} = \mathrm{Hom}(1_E, u), \bar{v} = \mathrm{Hom}(1_E, v)$) be exact.

Suppose that the sequence (5) is exact. Note first that

(II, $§ 1$, no. 2, formula (10)) since $v \circ u = 0$. The image of $\mathrm{Hom}(E, F')$ under $\bar{u}$ is therefore contained in the kernel $N$ of $\bar{v}$; let $f$ be the homomorphism of the $\mathbf{Z}$-module $\mathrm{Hom}(E, F')$ into $N$ whose graph is equal to that of $\bar{u}$; it is necessary to prove that $f$ is bijective and hence to define a mapping $g: N \to \mathrm{Hom}(E, F')$ such that $f \circ g$ and $g \circ f$ are the identity mappings. For this, let $w$ be an element of $N$, that is a linear mapping $w: E \to F$ such that $v \circ w = 0$. The latter relation is equivalent to $w(E) \subset \mathrm{Ker}(v) = u(F')$ by hypothesis, hence, since $u$ is injective, there exists one and only one linear mapping $w': E \to F'$ such that $w = u \circ w'$ and we take $g(w) = w'$; it is immediately verified that $g$ satisfies the desired conditions.

Conversely, suppose that sequence (6) is exact for every $A$-module $E$. As $\mathrm{Hom}(1_E, v \circ u) = \bar{v} \circ \bar{u} = 0$, then $v \circ u \circ w = 0$ for every homomorphism $w: E \to F'$. Taking $E = F'$ and $w = 1_{F'}$, it is seen first that $v \circ u = 0$ and hence $u(F') \subset \mathrm{Ker}(v)$. Now we take $E = \mathrm{Ker}(v)$ and let $\phi: E \to F$ be the canonical injection. Then $\bar{v}(\phi) = v \circ \phi = 0$ by definition and hence there exists $\psi \in \mathrm{Hom}(E, F')$ such that $\phi = \bar{u}(\psi) = u \circ \psi$, which obviously implies $\mathrm{Ker}(v) \subset u(F')$ and completes the proof of the exactness of (5) at $F$. Finally, if $\theta$ is the identity mapping of $\mathrm{Ker}\, u$, then $\bar{u}(\theta) = 0$, hence $\theta = 0$ and $\mathrm{Ker}\, u = \{0\}$, which proves the exactness of (5) at $F'$.

It suffices to apply Theorem 2 to the case where $F' = \{0\}$.

Starting from an exact sequence

the corresponding sequence

is not necessarily exact, in other words $\bar{v}$ is not necessarily surjective. If $F'$ is identified with a submodule of $F$ and $F''$ with the quotient module $F/F'$, this means that a linear mapping of $E$ into $F''$ is not necessarily of the form $v \circ w$, where $w$ is a linear mapping of $E$ into $F$. However:

(7)

splits (in other words, if $u(F')$ is a direct factor of $F$), the sequence

(8)

is exact and splits. Conversely, if sequence (8) is exact for every A-module E, the exact sequence (7) splits.

The first assertion follows from the fact that

is canonically identified with $\operatorname{Hom}(E, F' \oplus F'')$ by means of the $\mathbf{Z}$-linear mapping $\operatorname{Hom}(1_E, j') + \operatorname{Hom}(1_E, j''), j' : F' \to F' \oplus F''$ and $j'' : F'' \to F' \oplus F''$ being the canonical injections (§ 1, no. 6, Corollary 1 to Proposition 6). Conversely, if sequence (8) is exact for $E = F''$, there is an element $g \in \operatorname{Hom}(F'', F)$ such that $v \circ g = 1_{F''}$ and the conclusion follows from § 1, no. 9, Proposition 15.

Remark (2). The results of this no. are valid without modification for all commutative groups with operators.

For every A-module homomorphism u : E → F,

is identified with $\prod_i \operatorname{Hom}(1_{M_i}, u)$ (§ 1, no. 6, Corollary 1 to Proposition 6); the conclusion thus follows from Definition 1 and § 1, no. 5, Proposition 5 (ii).

It suffices by Proposition 3 to show that A_s is projective, which follows immediately from the commutativity of diagram (50) of § 1, no. 14.

(a) P is projective. (b) For every exact sequence 0 → F' → F → F'' → 0 of A-linear mappings, the sequence

is exact.

(c) For every surjective A-module homomorphism u : E → E'' and every homomorphism f : P → E'', there exists a homomorphism g : P → E such that f = u ∘ g (it is said that f can be "lifted" to a homomorphism of P into E).

(d) Every exact sequence 0 → E' → E → P → 0 of A-linear mappings splits (and therefore P is isomorphic to a direct factor of E).

(e) P is isomorphic to a direct factor of a free A-module.

It is trivial that (a) implies (b). To see that (b) implies (c), it suffices to apply (b) to the exact sequence 0 → E' → E → E'' → 0, where E' = Ker(u), since (c) expresses the fact that

is surjective. To see that (c) implies (d), it suffices to apply (c) to the surjective homomorphism $v : E \to P$ and the homomorphism $l_P : P \to P$; the existence of a homomorphism $g : P \to E$ such that $l_P = v \circ g$ implies that the sequence

splits (§ 1, no. 9, Proposition 15). As for every A-module M there exist a free A-module L and an exact sequence $0 \to R \to L \to M \to 0$ (§ 1, no. 11, Proposition 20), clearly (d) implies (e). Finally (e) implies (a) by virtue of Proposition 3 and its Corollary.

The condition is obviously sufficient; conversely, a finitely generated projective module E is isomorphic to a quotient of a free module F with a finite basis (§ 1, no. 11) and E is isomorphic to a direct factor of F by virtue of Proposition 4 (d).

It may be assumed that there are two finitely generated free C-modules M, N such that $M = E \oplus E', N = F \oplus F'$; it follows from § 1, no. 6, Corollary 1 to Proposition 6 that $\operatorname{Hom}_C(M, N)$ is finitely generated and free and on the other hand that $\operatorname{Hom}_C(M, N)$ is isomorphic to

whence the corollary.

It is the particular case of the canonical isomorphism $E \to \mathrm{Hom}_A(A_s, E)$ of § 1, no. 14, Remark 2, corresponding to $E = A_s$; the commutativity of diagram (50) of § 1, no. 14, shows that we have here an isomorphism of right A-modules.

If $A_d$ is identified with the dual of $A_s$ by means of the isomorphism of Proposition 5, the canonical bilinear form on $A_s \times A_d$ is then expressed by

Similarly, the dual of $A_d$ is canonically identified with $A_s$, the canonical bilinear form on $A_d \times A_s$ being expressed by

These results are immediate consequences of the definitions.

There is an analogous proposition (which we shall leave to the reader to state) for submodules of E orthogonal to subsets of E*.

If M is a submodule of E, M' the submodule of E* orthogonal to M and M'' the submodule of E orthogonal to M', then M ⊂ M'' but it may be that M ≠ M'' (Exercise 9). Note however that if M'' is the orthogonal of M'' in E*, then M'' = M'; for M' ⊂ M'' and on the other hand the relation M ⊂ M'' implies M'' ⊂ M'.

This follows immediately from (15).

The orthogonal of E in $E^*$ is 0.

If $u : E \to F$ is an isomorphism, $^t u : F^* \to E^*$ is an isomorphism and if $v : F \to E$ is the inverse isomorphism of $u$, $^t v$ is the inverse isomorphism of $^t u$ (formulae (17) and (18)).

an exact sequence of linear mappings. Then the sequence of transpose mappings

is exact.

If $j : M \to E$ is the canonical injection, the kernel of ${}^t j$ is by definition the orthogonal of $M$ in $E^*$.

Moreover, in the notation of the corollary, an injective homomorphism $E^*/M' \to M^*$ is obtained from ${}^t j$ when passing to the quotient.

is an isomorphism of the dual $E^*$ of $E$ onto the product $\prod_{i \in I} E_i^*$.

This is a particular case of § 1, no. 6, Corollary 1 to Proposition 6, applied to the case where $\prod_{\lambda} F_{\lambda} = A_s$.

If, by means of the canonical injections $j_i$, the $E_i$ are identified with submodules of their direct sum $E$ and if, by means of the product mapping $x^* \mapsto (tj_i(x^*))$, $E^*$ is identified with $\prod_{i \in I} E_i^*$, it can then be said that $\prod_{i \in I} E_i^*$ is the dual of $\bigoplus_{i \in I} E_i$, the canonical bilinear form being given by

(22)

$p \circ i = 1_M, \quad q \circ j = 1_N, \quad p \circ j = q \circ i = 0, \quad i \circ p + j \circ q = 1_E$, whence, by transposition (no. 5, formulae (16), (17) and (18)), $^ti \circ ^tp = 1_{M^*}, \ ^tj \circ ^tq = 1_{N^*}, \ ^tj \circ ^tp = ^ti \circ ^tq = 0. \ ^tp \circ ^ti + ^tq \circ ^tj = 1_{E^*}$ and the proposition follows from § 1, no. 6, Corollary 2 to Proposition 6.

Under the hypotheses of the Corollary, $M^*$ (resp. $N^*$) is often identified with the orthogonal $^tp(M^*)$ (resp. $^tq(N^*)$) of $N$ (resp. $M$) in $E^*$, thus identifying every linear form $u$ on $M$ (resp. $N$) with the linear form on $E$ extending $u$ and which is zero on $N$ (resp. $M$).

When an A-module $E$ admits a basis $(e_t)_{t \in T}$, it has been seen that giving this basis defines canonically an isomorphism $u : A_s^{(T)} \to E$. By virtue of Proposition 10 and no. 3, Proposition 5, the dual of $A_s^{(T)}$ is canonically identified with the product $A_d^T$; consider the contragredient isomorphism $\tilde{u} : A_d^T \to E^*$. If, for all $t \in T, f_t$ is the element of $A_d^T$ all of whose projections are zero with the exception of that of index $t$, which is equal to 1, and if we write $e_t^* = \tilde{u}(f_t)$, the elements $e_t^*$ of $E^*$ are, by (19) and (22), characterized by the relations

(23)

It amounts to the same to say that, for all $x = \sum_{t \in T} \xi_t e_t \in E, \ e_t^*(x) = \xi_t$; also $e_t^*$ is called the coordinate form of index $t$ on $E$. It follows from (23) that $(e_t^*)$ is a free system in $E^*$.

In particular, if $T$ is finite, the $e_t^*$ form a basis of $E^*$, the $f_t$ then forming the canonical basis of $A_d^T$. Hence:

Note that the dual of a free module with an infinite basis is not necessarily a free module (VII, § 3, Exercise 10).

A finitely generated projective left A-module can be identified with a direct factor M of a free A-module $A_s^n$ with a finite basis (no. 2, Corollary 1 to Proposition 4). Then (Proposition 11 and Corollary to Proposition 10) M* is isomorphic to a direct factor of $A_d^n$, whence the corollary.

There exists a surjective homomorphism $u : L \to E$, where $L = A_s^{(T)}$, such that if $(e_t)_{t \in T}$ is the canonical basis of L then $u(e_t) = a_t$ (§ 1, no. 11, Proposition 17); for E to be projective, it is necessary and sufficient that there exist a linear mapping $v : E \to L$ such that $u \circ v = 1_E$ (no. 2, Proposition 4 and § 1, no. 9, Proposition 15). If such a mapping exists and we write ${}^tv(e_t^*) = a_t^*$, then $\langle x, a_t^* \rangle = \langle x, {}^tv(e_t^*) \rangle = \langle v(x), e_t^* \rangle$, hence the family $(\langle x, a_t^* \rangle)$ has finite support and $x = u \left( \sum_{t \in T} \langle (x), e_t^* \rangle e_t \right) = \sum_{t \in T} \langle x, a_t^* \rangle a_t$ for all $x \in E$. Conversely, if condition (b) of the statement is fulfilled, the sum $\sum_{t \in T} \langle x, a_t^* \rangle e_t$ is defined for all $x \in E$ and $x \to \sum_{t \in T} \langle x, a_t^* \rangle e_t$ is a linear mapping $v : E \to L$ such that $u \circ v = 1_E$.

Let $(e_t)_{t \in T}$ be a basis of E and let $(e_t^*)$ be the family of corresponding coordinate forms; by definition, if $x \in E$ is such that $\tilde{x} = 0$, then $\langle x, e_t^* \rangle = 0$ for all $t \in T$, in other words all the coordinates of $x$ are zero, hence $x = 0$. Suppose further that T is finite; since $\langle \tilde{e}_t, e_{t'}^* \rangle = \delta_{tt'}$, $(\tilde{e}_t)$ is the dual basis of $(e_t^*)$ in $E^{**}$ and, as $c_E$ transforms a basis of E into a basis of $E^{**}$, $c_E$ is bijective (§ 1, no. 11, Corollary 3 to Proposition 17). We have moreover proved: