Continuity question on topological product space

Question

Let $\left\{X_a \right\}_{a \in I}$ be an indexed family of topological spaces. Consider their topological product $X=\prod_{a \in I} X_a$. Let $I'$ be a finite subset of $I$ and define $X' =\prod_{a \in I'} X_a$. Let $y$ be a point of $X$. Define an embedding $j:X' \rightarrow X$ by $j(x')(a)=x'(a)$ if $a \in I'$ and $j(x')(a)=y(a)$ if $a \in I-I'$, where $x' \in X'$. Let $p_a : X \rightarrow X_a$ be the projection on the $a$ factor $x \mapsto x(a)$.

I read the statement "j is continuous, since all maps $p_a j$ are continuous". I can see that $p_a j$ is continuous for any $a \in I$. But why does this continuity imply the continuity of $j$?

Answer 1

It's a special case of a general result that's worth knowing:

Theorem: Let $X=\prod\limits_{i\in I}X_i$ be a product of topological spaces, and for $i\in I$ let $p_i:X\to X_i$ be the canonical projection map. Let $Y$ be a topological space, and let $f:Y\to X$ be any function. Then $f$ is continuous iff $p_i\circ f:Y\to X_i$ is continuous for each $i\in I$.

Clearly continuity of $f$ implies continuity of the compositions $p_i\circ f$, since the composition of continuous functions is continuous. The converse implication is an easy consequence of the fact that the sets of the form

$$\bigcap_{i\in F}p_i^{-1}[U_i]\;,$$ where $F$ is a finite subset of $I$, and $U_i$ is open in $X_i$ for each $i\in F$, are a base for the product topology.

Answer 2

Since each $p_a j$ is continuous, this means that for each $a$ and each open $U \subset X_a$, we have $j^{-1}(p_a^{-1}(U))$ is open in $X'$. Now note that the collection of sets of the form $p_a^{-1}(U)$ generate the topology of $X$ (just by definition of the product topology).