The first time I come across the functions defined on Cartesian product is when I read this:
As indicated in the above picture, the facts 2 and 3 are straightforward and I can prove it.
However, when I learned the homotopy theory, I encountered a more general version of this kind of functions:
Now I will use the notations shown in the second image.
Regarding some proofs in the homotopy theory, I was really stuck. For example, let me try to prove "$f$ is homotopic to $f$ itself". Books say that it is trivial that $ f \simeq f$, the map $F(x, t) = f (x)$ is the required homotopy. But I think is not trivial, $F(x,t)$ is a map on product space, $f(x)$ is a map on a single place $X$. Even if $f(x)$ is continuous, and their values are equal, how could the fact that $F(x,t)$ is continuous so obvious? In my mind, they are totally different things. I can only prove $F(x,t)$ is continuous when $t$ is fixed at $t_0$ as follows:
Since $f$ is continuous from $X$ to $Y$, so for any open set $U$ in Y , there is an open set $V$ such that $V = f^{-1}(X)$ in $X$. since V is open in a subspace X, which is an open subspace of the whole space $X \times I$, so V is also open in $X \times I$, which conclude that $F(x,t_0)$ is continuous.
Then, I can't reason further.
I am wondering if I have missed some basic multivariable calculus concepts so that I am stuck on such a trivial problem. But I can't find any book dealing directly with continuity about functions defined on Cartesian products. I have been self-learning for a long while without supervision, so maybe this question is really silly. I'd really appreciate it if anyone could give me some references so I can read them and continue learning.
EDIT: I've just found two ICM papers about this: http://matwbn.icm.edu.pl/ksiazki/fm/fm59/fm59123.pdf http://matwbn.icm.edu.pl/ksiazki/fm/fm106/fm10618.pdf
It's indeed considered trivial.
The projection map $\pi_X:X\times I \to X$ is continuous, and $F=f\circ \pi_X$, which is thus also continuous.
You can also prove it directly from the definition: if $U\subseteq Y$ is open, then $$F^{-1}(U)=f^{-1}(U)\times I$$ which is an open set in the product topology on $X\times I$, since $f^{-1}(U)$ is open.