I have the following problem:
Leg $I$ be a nonempty index set and let $(M_i,\mathfrak{M_i})$ and $(N_1,\mathfrak{N_i})$ be topological spaces. Moreover let $f_i:M_i\rightarrow N_i$ be maps. Finally endow $M=\prod_{i\in I} M_i$ and $N=\prod_{i\in I} N_i$ with the product topology. Show that the product map $$f:M\rightarrow N$$ is continuous iff all maps $f_i$ are continuous.
My idea was the following:
$\Rightarrow $ Let us first assume that the product map is continous. Then define $p_i:M\rightarrow M_i$ and $q_i:N\rightarrow N_i$ be the projection maps which are continous since $M,N$ are endowed with the product topology. Then we see that we have $$f_i\circ p_i=q_i\circ f$$ and remark that $q_i\circ f$ is the composition of two continuous functions. Therefore we can deduce that $f_i\circ p_i$ is continuous. But now it is clear that $f_i$ needs to be continuous, otherwise we would have an open set $O\subset N_i$ such that $f_i^{-1}(O)$ is not open in $M_i$ and therefore $p_i^{-1}(f_i^{-1}(O))$ is not open in $M$ but this would contradict the fact that $f_i\circ p_i$ is continuous.
$\Leftarrow$ I would have done a similar argument for $f$ in the other direction.
Now I'm not sure if this works. Could someone help me maybe?
Thank you!
You need the assumption that all $M_i$ are non-empty. Otherwise the product $M$ is empty and you cannot get any information about the $f_i$.
Your proof of $\Rightarrow$ is correct.
For $\Leftarrow$ you can argue as follows:
If all $f_i$ are continuous, then all $f_i \circ p_i : M \to N_i$ are continuous. Thus all $q_i \circ f = f_i \circ p_i$ are continuous. The universal property of the product $N$ shows then that $f$ is continuous.