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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

Aryaman Maithani
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user123234
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  • Can we assume that the $M_i$ and $N_i$ are nonempty? Otherwise, we may face issues. – Aryaman Maithani Oct 19 '21 at 08:21
  • oh yes they are non empty. Does then my proof holds in this case? – user123234 Oct 19 '21 at 08:23
  • Suppose that one $f_j$ were not a homeomorphism then we'd have an immediate contradiction. The other direction is pretty trivial. – Henno Brandsma Oct 19 '21 at 11:42
  • What theorem allows you to conclude that $f$ continuous implies all $f_i$ continuous? – Henno Brandsma Oct 19 '21 at 11:43
  • we proofed this statement in ex. a) this is now ex b) – user123234 Oct 19 '21 at 11:44
  • So as I understand you, to show $\Leftarrow$ you do it by contradiction i.e. assumung that one $f_j$ is no homeomorphism but that $f$ is one right? – user123234 Oct 19 '21 at 11:46
  • Assuming $f$ is one, all $f_i$ must be is the question, so assume one is not and derive a contradiction. – Henno Brandsma Oct 19 '21 at 11:59
  • The bijectivity is trivial both ways. – Henno Brandsma Oct 19 '21 at 12:00
  • @HennoBrandsma sorry this is maybe a bit late but I found this question really interesting and wanted to ask something. you spoke about a contradiction. If I assume that $f$ is a homeomorphism and we have one $f_j$ which is not a homeomorphism, then since $f_i\circ q_i=p_i\circ f$ and $p_i\circ f$ is continuous but $f_i\circ q_i$ doesn't that here we have a contradiction? – user1294729 Jan 27 '22 at 22:54

2 Answers2

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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.

Henno Brandsma
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  • sorry this is maybe a bit late but I found this question really interesting and wanted to ask something. you spoke about a contradiction. If I assume that $f$ is a homeomorphism and we have one $f_j$ which is not a homeomorphism, then since $f_i\circ q_i=p_i\circ f$ and $p_i\circ f$ is continuous but $f_i\circ q_i$ doesn't that here we have a contradiction? – user1294729 Jan 27 '22 at 22:56
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$\newcommand{\m}{\mathbf{m}}$As commented, I'll be assuming that all the spaces (including the product spaces) are nonempty.
In particular, I'm fixing two points $\m = (m_i)_i \in M$ and $\mathbf{n} = (n_i)_i \in N$ in the product spaces.

($\Rightarrow$)

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.

This needs a little argument. You would need to show that any open set in $M_i$ can be written as $q_i(U)$ for some open $U \subset M$. Since this is true, it does follow that $f_i$ is open.

Now, you want to show that each $f_i$ is injective. Fix $j \in I$ and let $x_1, x_2 \in M_{j}$ be arbitrary. Suppose $f_{j}(x_1) = f_{j}(x_2)$. Now, consider the elements $\m_1, \m_2 \in M$ which are equal to $\m$ in all coordinates but the $j$-th one, and $(\m_1)_j = x_1$ and $(\m_2)_j = x_2$.
Then, $f(\m_1) = f(\m_2)$. Since $f$ is injective, $\m_1 = \m_2$ and hence, $x_1 = x_2$.

Similarly, to show that $f_j$ is surjective, pick $y \in N_j$ and define $\mathbf{n}' \in N$ to be equal to $\mathbf{n}$ in all coordinates but the $j$-th one and...

Can you try the other direction now?

Added. How do we see that $f_i$ are open?

Fix $j \in I$. Let $V \subset M_j$ be an arbitrary open set. We wish to show that $f_j[V]$ is open. But note that $V = q_j[U]$, where $U$ is the subset of $M$ defined to be the product of all the $M_i$ except that in the $j$-th coordinate, it is $V$.

Thus, we have $$f_j[V] = (f_j \circ q_j)[U] = (p_j \circ f)[U],$$ which is open since $p_j$ and $f$ are open maps, and $U$ is open (it is a basis element of the product topology).

Aryaman Maithani
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  • sorry I stuck at the first part, i.e. showing that $f_i$ is open. Does we need the basis of the subspace topology? – user123234 Oct 19 '21 at 09:08
  • Another small question about bijectivity. Can't we argue like this: Since $f$ is bijective it is also injective and we know that $p_i$ is surjective. Therefore it follows that $p_i\circ f=f_i\circ q_i$ is bijective. But since $q_i$ is surjective the only possibility is that $f_i$ is injective. In my opinion I now only need to show the surjectivity of $f_i$, don't I? – user123234 Oct 19 '21 at 09:19
  • @Strandkorb: Hi, do you still need these clarified? – Aryaman Maithani Oct 19 '21 at 14:27
  • Hi, yes if you have time sure, because then I see several possibilities to solve this! – user123234 Oct 19 '21 at 15:40
  • @Strandkorb: I have added details about showing $f_i$ is open. I do use the basis of the product (not subspace) topology. – Aryaman Maithani Oct 19 '21 at 15:46
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    As for your second comment, let us look at the general statement: Suppose $f, g$ are general functions such that $f \circ g$ is a bijection. Then, you can conclude that $f$ is a surjection and $g$ is an injection. Now, if you are also given that $g$ is a surjection, then $g$ will turn out to be a bijection as well. In which case, you also get that $f = (f \circ g) \circ g^{-1}$ is a bijection and you are done directly. – Aryaman Maithani Oct 19 '21 at 15:51
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    But I think that you have an error in your second comment and you want to say that "we know $f_i \circ q_i$ is surjective and $q_i$ is also surjective". From this also we can conclude that $f_i$ is surjective, yes. (We don't even need $q_i$ to be surjective, just the composition being surjective is enough.) – Aryaman Maithani Oct 19 '21 at 15:53
  • Sorry I don't see why we can find such a U? could you please explain this? Because our definition of the product topology is, that it is determined by the subbasis $S={p_i^{-1}(O_i)|i\in I, O_i\in M_i}$ where $p_i:M\rightarrow M_i$ are the projections. But then I understood it that an element of M can be written as a intersection of finite unions of elements in S. And this is somehow different to what you claimed or not? – user123234 Oct 19 '21 at 17:25
  • @Strandkorb: I explicitly defined $U$ to be the product of $M_i$ except that we have $V$ instead of $M_j$. In your notation, we have $U = p_j^{-1}[V]$. Since $p_j$ is surjective, we have $V = p_j[p_j^{-1}[V]] = p_j[U]$. – Aryaman Maithani Oct 19 '21 at 17:42
  • and why can I chose $U=p_j'{-1}(V)$ – user123234 Oct 19 '21 at 18:01
  • @Strandkorb: What do you mean why? – Aryaman Maithani Oct 19 '21 at 18:04
  • sorry I wantet to write wha can I chose $U=p_j^{-1}(V)$ – user123234 Oct 19 '21 at 18:25
  • @Strandkorb: because you can – Aryaman Maithani Oct 19 '21 at 20:10
  • Ah okey don't we need any condition to be satisfied – user123234 Oct 19 '21 at 20:15
  • @Strandkorb: No, I wanted to show that $V = p_j[U]$ for some open $U$ and then I did exactly that by constructing an explicit candidate. – Aryaman Maithani Oct 19 '21 at 20:48
  • ah okey makes sense thanks – user123234 Oct 20 '21 at 17:05