I have the following basic question regarding continuous functions.
Is the following statement correct?
Fix nonempty toplogical spaces $A, B, C, D$ with $A \times C$ and $B \times D$ endowed with the product topology. Also, fix functions
\begin{align*} h & : A \times C \to B \times D ,\\ f &: A \to B ,\\ g &: C \to D , \end{align*}
with $h := (f, g)$. Thus, if $h$ and $f$ are continuous, then $g$ is continuous.
I know that if $f$ and $g$ are continuous, then $h$ is continuous, while if $h$ is continuous it is not necessarily the case that both $f$ and $g$ are continuous. What about this 'intermediate' case?
Thanks a lot in advance for any feedback.
The function $h$ is continuous if and only if $f,g$ are.
For one direction, assume $f,g$ are continuous and pick a net $(x_\alpha,y_\alpha)_\alpha$ in $A\times C$ converging to $(x,y)$. Then $\lim_\alpha x_\alpha=x$ in $A$ and $\lim_\alpha f(x_\alpha)=f(x)$ by continuity of $f$. Similarly, $\lim_\alpha g(y_\alpha)=y$. So $\lim_\alpha h((x_\alpha, y_\alpha))=(x,y)$, and $h$ is continuous.
For the other direction, assume $h$ is continuous. Fix any $y\in C$ (here is where the non-emptiness of $C$ becomes necessary). Define $a:A\to A\times C$ by $a(x)=(x,y)$. Define $b:B\times D\to B$ by $b(u,v)=u$. Then $a,b$ are continuous and $f=b\circ h\circ a$. Thus $f$ is also continuous, since it is the composition of continuous functions. By symmetry, so is $g$.