Let $f : Z \to X \times Y$ such that $f (z) := ( f_1 (z) , f_2 (z))$, with $f_1 : Z \to X$, and $f_2 : Z \to Y$. In the following, $\tau_X$, $\tau_Y$, $\tau_{X \times Y}$, and $\tau_Z$ denote the topologies of the corresponding indexing sets.
Is the following a correct proof of the proposition?
Proposition: A function $f: Z \to X \times Y$ is continuous if and only if both its component functions $f_1$ and $f_2$ are continuous.
Proof: $[\Rightarrow]$ Let $f$ be continuous, and let $V \in \tau_X$, $U \in \tau_Y$. From the definition of product topology we have that $V \times U \in \tau_{X \times Y}$. From the continuity of $f$, $f^{-1} (V \times U) \in \tau_Z$. Let $A := f^{-1} (V \times U)$. Then, notice that from the definition of $f$ we have that $f_1 (A) = V$, that is $f^{-1}_1 (V) = A \in \tau_X$, with a similar argument holding for $U \in \tau_Y$.
$[\Leftarrow]$ Let $f_1 \in X^Z$, and $f_2 \in Y^Z$ be continuous. Let $G \in \tau_{X \times Y}$. Hence, from the definition of product topology there are open sets $V \in \tau_X$, and $U \in \tau_Y$ such that $G:= V \times U$. From the continuity of $f_1$, and $f_2$, we have that $f^{-1}_1 (V) \in \tau_Z$, and $f^{-1}_2 (U) \in \tau_Z$. Hence, from the definition of $f$, there is a $A \in \tau_Z$ such that $ A:= f^{-1}_1 (V) = f^{-1}_2 (U)$. $\square$
In particular I have doubts concerning the $[\Leftarrow]$ direction.
[Indeed I found [in this set of notes by Hatcher][1] that the argument used to prove the result there (which is much neater) is based on the fact that $$f^{-1} (U \times V) = f^{-1}_1 (U) \cap f^{-1}_2 (V),$$ which is a statement that is not completely clear to me.]
As always, any feedback is most welcome.
Thank you for your time.
Reference updated to new address. [1]: https://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf
If $x \in f^{-1}(U \times V)$, then $f(x)\in U \times V$, which implies $f_{1}(x) \in U$ and $f_{2}(x) \in V$, so $x \in f^{-1}_{1}(U)$ and $x \in f^{-1}_{2}(V)$. You can reverse these steps and then you have the equality of the sets.