First off, I am aware that there are many posts related to this theorem, but I believe my question is sufficiently different to warrant a seperate post.

Here is the problem.

Let $X,Y$ and $Z$ be metric spaces equipped with the "standard euclidean metric" (i.e. taking the square root of the metric squared and summed). Let $F: X \to Y \times Z, \ f_1: X \to Y, \ f_2:X \to Z$. Prove that $F$ is continuous iff $f_1$ and $ f_2$ are continuous.

There is a simple proof that relies on $X,Y$ and $Z$ being metric spaces. However I wanted to find a proof that didn't use any metrics etc. . Here is what I have so far:

Assume $f_1, f_2$ are continuous on $X$. Let $O \subset Y \times Z$ be open. Then, for some indexing set $I$, we can write $O = \cup_{i \in I} (U_i \times V_i)$, where the $U_i$ and $V_i$ are open in $Y$ and $Z$, respectively. Noting that preimages and unions commute we have,

$F^{-1}(O) = F^{-1}(\cup_{i \in I} (U_i \times V_i)) = \cup_{i \in I} F^{-1}(U_i \times V_i) = \cup_{i \in I} (f_1(U_i) \cap f_2(V_i))$ , which is open, as the intersection of a finite number of open sets (here two) is open and the arbitrary union of open sets is open. This proves $F$ is continuous.

This covers the first direction. But this is where I get stuck with a purely topological argument. Please note that I am fairly new to topology and know practically nothing of product topologies etc. ( this question was asked in my course on metric spaces).

If anyone could help me with this final direction ( in simple terms) that would be great! Many thanks.


I think if we consider $h : Y \times Z \to Y$ where $h(y,z) = y$, then we can get that $f_1$ is continuous because $f_1 (x) = h(F(x))$ (the composition of two continuous functions). $h$ is continuous, because $h^{-1} (U) = U \times Z$ for any open $U$ in $Y$.

And we can do the same thing for the projection onto $Z$.


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