a) If $f$ and $g$ are open (resp. closed), so is $g \circ f$.

b) If $g \circ f$ is open (resp. closed) and if $f$ is continuous and surjective, then $g$ is open (resp. closed).

c) If $g \circ f$ is open (resp. closed) and if $g$ is continuous and injective, then $f$ is open (resp. closed).

From Definition 1. a) follows immediately. To prove b) it is enough to remark that every open (resp. closed) subset $A'$ of $X'$ can be written as $f(A)$, where $A = f^{-1}(A')$ is open (resp. closed) in $X$ (§ 2, no. 1, Theorem 1); hence $g(A') = g(f(A))$ is open (resp. closed) in $X''$. Finally, to prove c), we remark that $f(A) = g^{-1}(g(f(A)))$ for every subset $A$ of $X$; by hypothesis, if $A$ is open (resp. closed) in $X$, then $g(f(A))$ is open (resp. closed) in $X''$, hence $f(A)$ is open (resp. closed) in $X'$ by § 2, no. 1, Theorem 1.

a) If $f$ is open (resp. closed), $f_T$ is open (resp. closed).

b) Let $(T(i))_{i \in I}$ be a family of subsets of $Y$ whose interiors cover $Y$, or which is a locally finite closed covering of $Y$; if all the $f_{T(i)}$ are open (resp. closed), then $f$ is open (resp. closed).

a) If $A$ is an open (resp. closed) subset of $f^{-1}(T)$, then there is an open (resp. closed) subset $B$ of $X$ such that $A = B \cap f^{-1}(T)$, and therefore f_T(A) = f(B) \cap T; by hypothesis, $f(B)$ is open (resp. closed), so that $f_T(A)$ is open (resp. closed) in $T$.

b) Let $B$ be an open (resp. closed) subset of $X$, and let $B_i$ denote $B \cap f^{-1}(T(i))$; then $f(B) \cap T(i) = f_{T(i)}(B_i)$. Since $f_{T(i)}(B_i)$ is open (resp. closed) in $T(i)$ by hypothesis, it follows that $f(B)$ is open (resp. closed) in $Y$, by Proposition 3 of § 3, no. 1.

For $f$ is clearly bijective, and is open by virtue of Proposition 2.

a) f is an open mapping.

b) The three mappings p, h, i are open.

c) The equivalence relation R is open, h is a homeomorphism, and $f(X)$ is an open subset of Y.

Also the preceding proposition remains true if "open" is replaced by "closed" throughout.

Since the injection i is continuous, it follows from Proposition 1 c) of no. 1 that if f is open then so is $h \circ p$; since p is surjective and continuous, Proposition 1 b) shows that h is open; h is in any case a continuous bijection; thus h is a homeomorphism, and therefore, from Proposition 1 a) of no. 1, $p = \overline{h}^{-1} \circ (h \circ p)$ is an open mapping. Also [no. 1, Proposition 1 b)] $i \circ h$ is open; hence [no. 1, Proposition 1 a)] so is $i = (i \circ h) \circ \overline{h}^{-1}$. This proves that a) implies b). Conversely, Proposition 1 a) of no. 1 shows that b) implies a). Finally, the equivalence of b) and c) follows immediately from the definitions.

The proof in the case of closed mappings is analogous, mutatis mutandis.

a) A is open (resp. closed) in X.

b) A is saturated with respect to R.

Then the relation $R_A$ induced on A is open (resp. closed) and the canonical mapping of $A/R_A$ onto $f(A)$ is a homeomorphism.

Consider the commutative diagram (i) of § 3, no. 6, which gives the canonical decomposition of $f \circ j$. Under condition a), $j$ is open (resp. closed) and so is $f$ by hypothesis; hence $f \circ j$ is open (resp. closed) [no. 1, Proposition 1 a)], and the result follows from Proposition 3. Under condition b) we have

and $h \circ \varphi$ is the mapping of $A$ into $f(A)$ which agrees with $f$ on $A$; by virtue of Proposition 2 a) of no. 1, $h \circ \varphi$ is open (resp. closed), and the result again follows from Proposition 3, applied to $h \circ \varphi$.

a) $f$ is an open mapping.

b) For each $U \in \mathfrak{B}$, $f(U)$ is open in $Y$.

c) For each $x \in X$ and each neighbourhood $V$ of $x$ in $X$, $f(V)$ is a neighbourhood of $f(x)$ in $Y$.

The equivalence of a) and b) follows immediately from the definitions and from $(O_1)$; the equivalence of a) and c) is a consequence of Proposition 1 of § 1, no. 2.

a) The relation $R$ is open.

b) The interior of each subset which is saturated with respect to $R$ is saturated with respect to $R$.

c) The closure of each subset which is saturated with respect to $R$ is saturated with respect to $R$.

By taking complements ($§ 1$, no. 6, formula (2)) we see that b) and c) are equivalent. Let us show that b) implies a): suppose condition b) is satisfied and let $U$ be an open subset of $X$, $V$ its saturation with respect to $R$; then $\dot{V} \supset U$, and since by hypothesis $\dot{V}$ is saturated, it follows that $\dot{V} = V$, and therefore the saturation of $U$ is open. Conversely, suppose condition a) is satisfied, and let $A$ be a saturated set; if $B$ is the saturation of $\dot{A}$, then $\dot{A} \subset B \subset A$, and since $B$ is open by hypothesis it follows that $B = \dot{A}$.

Each of the two assertions of the proposition can be deduced from the other by taking complements and using formula (2) of § 1, no. 6 and the fact that if $B$ is a saturated subset of $X$ then $\varphi(\complement B) = \complement_{\varphi}(B)$. By virtue of Proposition 6, $\overline{A}$ is saturated; hence $\varphi(\overline{A})$ is closed in $X/R$, and since $A \subset \overline{A}$ we have $\varphi(A) \subset \varphi(\overline{A})$, so that $\varphi(A) \subset \varphi(\overline{A})$. But since $\varphi$ is continuous, $\varphi(\overline{A}) \subset \varphi(A)$ (§ 2, no. 1, Theorem 1), and the result follows.

By virtue of Proposition 5 we need only prove that the image under $f$ of any elementary set $\prod_{i \in I} A_i$ in $\prod_{i \in I} X_i$ is open in $\prod_{i \in I} Y_i$. But this image is $\prod_{i \in I} f_i(A_i)$, and the hypotheses imply that $f_i(A_i)$ is open in $Y_i$ for each $i \in I$, and that $f_i(A_i) = Y_i$ for all but a finite number of indices; whence the result.

"for each $i \in I$, $\operatorname{pr}_i(x) = \operatorname{pr}_i(y)$ (mod $R_i$)"

between $x$ and $y$, and let $f$ be the product mapping $(x_i) \to (f_i(x_i))$ of $X$ into $\prod_{i \in I} (X_i/R_i)$. If each of the relations $R_i$ is open, then the relation $R$ is open, and the bijection associated with $f$ is a homeomorphism of $X/R$ onto $\prod_{i \in I} (X_i/R_i)$.

$R$ is the relation $f(x) = f(y)$. Since $f$ is continuous and open by Proposition 8 above and § 4, no. 1, Corollary 1 of Proposition 1, the result follows from Proposition 3 of no. 2.

In particular, if $R$ (resp. $S$) is an open equivalence relation on a topological space $X$ (resp. $Y$), then the canonical bijection of

is a homeomorphism. If R and S are not assumed to be open, this bijection is continuous but is not necessarily a homeomorphism, even if one of the relations R, S is the relation of equality (Exercise 6).

The condition is sufficient, for it obviously implies that f is closed, and it also implies that f is continuous by reason of § 2, no. 1, Theorem 1. Conversely, if f is continuous and closed, we have f(A) ⊂ f(Ā) ⊂ f(A) by § 2, no. 1, Theorem 1; also f(Ā) is closed in X' by hypothesis; hence f(Ā) = f(A).

Suppose R is closed, and let U be an arbitrary open neighbourhood of M; since F = ĈU is closed in X, the saturation S of F with respect to R is closed in X. Since M is saturated with respect to R, we have M ∩ S = ∅, and thus V = ĈS is an open neighbourhood of M, saturated with respect to R and contained in U.

To prove the converse, let F be any closed subset of X. Let T be the saturation of F with respect to R, let x be a point of ĈT, and let M be the equivalence class of x; then M ∩ T = ∅ and a fortiori M ∩ F = ∅, so that U = ĈF is a neighbourhood of M. Hence there is a neighbourhood V ⊂ U of M such that V is saturated with respect to R; V does not meet F, hence does not meet T, so that ĈT is a neighbourhood of M and therefore of x. This shows that ĈT is open (§ 1, no. 2, Proposition 1), i.e. T is closed.