Proposition 4

Let $I$ be a finite set and for each $i \in I$ let $f_i : X_i \to Y_i$ be a continuous mapping. Let $X = \prod_{i \in I} X_i, \ Y = \prod_{i \in I} Y_i,$ and let $f : X \to Y$ be the product mapping $(x_i) \to (f_i(x_i))$. Then: a) If each of the $f_i$ is proper, then $f$ is proper. b) If $f$ is proper and if the $X_i$ are non-empty, then each of the $f_i$ is proper.

(We shall see in no. 2, Theorem 1, Corollary 3, that this proposition extends to infinite products.)

By induction it is enough to consider the case where $I = \{1, 2\}$. a) Suppose that $f_1, f_2$ are proper, and let $Z$ be a topological space; $f_1 \times f_2 \times \iota_Z$ is the composition of $\iota_{Y_1} \times f_2 \times \iota_Z$ and

these two mappings are closed by hypothesis, hence so is $f_1 \times f_2 \times \iota_Z$ [§ 5, no. 1, Proposition 1 a)], whence $f_1 \times f_2$ is proper.

b) Now suppose $f$ is proper. Let $F$ be a closed subset of $X_2 \times Z$ and let $G$ be the image of $F$ in $Y_2 \times Z$ under the mapping $f_2 \times \iota_Z$. Then the image of $X_1 \times F$ in $Y_1 \times Y_2 \times Z$ under $f_1 \times f_2 \times \iota_Z$ is $f_1(X_1) \times G$. By hypothesis, this is closed in $Y_1 \times Y_2 \times Z$; if $X_1 \neq \emptyset$, then $f_1(X_1)$ is not empty, which implies that $G$ is closed in $Y_2 \times Z$ (§ 4, no. 3, Corollary to Proposition 7); hence $f_2$ is proper. Similarly $f_1$ is proper if $X_2 \neq \emptyset$.