I have the following problem:
Let $I$ be a nonempty index set and let $(M_i,T_{M_i})$ and $(N_i,T_{N_i})$ 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 map $f=\prod_{i\in I} f_i: M\rightarrow N$ is a homeomorphism iff each $f_i$ is a homeomorphism.
Proof
We have just shown that $f$ is continuous iff each $f_i$ is continuous. From the lecture we know that a map $f$ between top. spaces is a homeomorphism iff $f$ is continuous, bijective and open.
$\Rightarrow$ Let us assume that $f$ is a homeomorphism. We need to show that all $f_i$ are continuous, bijective and open. Since $f$ is continuous, we can immediately deduce that all $f_i$ are continuous. Let us denote $p_i:N\rightarrow N_i$ and $q_i:M\rightarrow M_i$ the projection maps. Since $M,N$ are endowed with the product topology we know that $p_i,q_i$ are both open. Thus we get immediately that $p_i\circ f$ is open. But since $p_i\circ f=f_i\circ q_i$ and $q_i$ is open, we can also deduce that $f_i$ has to be open. Now we only need to show that $f_i$ are bijective. We know that the projection map is surjective.
Does it is correct till this point or do I wrote nonsense? (I would do the other inclusion similarly using the fact that $f_i$ are continuous, bijective and open)
But for the bijectivity I have some problems. Could one gave me a hint?
Thank you
Best Answer
If $f$ is bijective assume some $f_j$ is not. Either $f_j$ is not injective and so for some $x \neq y \in M_j$ we have $f_j(x)=f_j(y)$. Add dummy points for all other coordinates and note that "extended" $x$ and $y$ also are mapped by $f$ to the same value, a contradiction. If $f_j$ is not surjective some $q\in N_j$ is not in $f_j[M_j]$ but then stuffing $y$ to a point of $N$ (so we're in both cases making the essential assumption that $M,N \neq \emptyset$ etc.) it also cannot be in $f[M]$, contradiction.
The bijective part is easiest in a way (assuming non-empty products).
If $f_j$ were not open, then for some open $O \subseteq M_j$ we'd have a non-open image, but then $f[\pi_j^{-1}[O]]$ is non-open as $\pi_j[f[\pi_j^{-1}[O]]]= f_j[O]$ is non-open and projections are open maps. (I use the same $\pi_j$ notation for projections on both $M$ and $N$). Continuity can be shown the same way:
$\pi_j[f^{-1}[\pi_j^{-1}[O]]]=f_j^{-1}[O]$ so a non-open pre-image for some $f_j$ would also give one for $f$ etc.