# $\coprod_{i \in I} (Y \times X_{i})$ and $Y \times \coprod_{i \in I}X_{i}$ are homeomorphic

To prove that $\coprod_{i \in I} (Y \times X_{i})$ and $Y \times \coprod_{i \in I}X_{i}$ are homeomorphic, I constructed $h: \coprod_{i \in I} (Y \times X_{i}) \rightarrow Y \times \coprod_{i \in I}X_{i}: ((y,x),i) \longmapsto (y,(x,i)).$

Now it is clear without any proof that $h$ is bijective.

To prove that $h$ is continuous I did the following:

$A \subseteq Y \times \coprod X_{i}$ is open $\iff A = B \times C, B \subseteq Y$ open, $C \subseteq \coprod X_{i}$ open $\iff A = B \times C, B \subseteq Y$ open, $C \cap X_{i}$ open in $X_{i}, \forall i \in I$.

To prove that $h^{-1}(A)$ is open in $\coprod Y \times X_{i}$, we must show that $A \cap (Y \times X_{i})$ is open in $Y \times X_{i}, \forall i \in I$. Now we have $A \cap (Y \times X_{i}) = (B \times C) \cap (Y \times X_{i}) = (B \cap Y) \times (C \cap X_{i}) = B \times (C \cap X_{i})$, where the first component is open in $Y$ and the second in $X_{i}$ because of the previous facts.

Is this reasoning correct?

For the second part of the prove, we must show that $h^{-1}$ is continuous, or, equivalent, that $h$ is open. Therefore I take $A \subseteq \coprod (Y \times X_{i})$ open. This means that $\forall i \in I: A \cap (Y \times X_{i})$ open in $Y \times X_{i}$. So write $A = E \times F$, to prove that $h(A)$ is open in $Y \times \coprod X_{i}$ we must show that $E \cap Y$ is open in $Y$ (which is already the case) and that $\forall i \in I F\cap X_{i}$ is open in $X_{i}$ (which is also true).

Is this second reasoning also correct?

If not, where do I go wrong?

As always, thanks for the help!

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An open subset $A \subset X_1\times X_2$ is not generically of the form $A=O_1\times O_2$, with $O_1$ and $O_2$ opens of $X_1$ and $X_2$ respectively. Think about the (open) unit disc in $\mathbb R^2$. –  Student May 1 '12 at 11:03
I formulated that quite badly... The thing is, that for continuous maps, you only need to check what happens with the basis for the topology. And if $X$ and $Y$ are topological spaces, a basis of the product topology on $X \times Y$ is given by $U \times V$, with $U$ open in $X$ and $V$ open in $Y$. –  KarenVO May 1 '12 at 11:11

Define $h: \coprod (Y \times X_{i}) \to Y \times \coprod X_{i}: ((y,x),i) \to (y, (x,i))$.

It is clear that $h$ is bijective.

We now prove that $h$ is continuous: Consider an open set in the basis for the product topology on $Y \times \coprod X_{i}$. This set is of the form $E \times F$, with $E$ open in $Y$, $F$ open in $\coprod X_{i}$.

$F$ is open in $\coprod X_{i}$ if and only if $F \cap X_{i}$ is open in $X_{i}$, for every $i \in I$.

Then we have:

$h^{-1}(E \times F)$ open in $\coprod(Y \times X_{i})$

$\Leftrightarrow h^{-1}(E \times F) \cap (Y \times X_{i})$ open in $Y \times X_{i}$, $\forall i \in I$

$\Leftrightarrow (E \times F) \cap (Y \times X_{i})$ open in $Y \times X_{i}$, $\forall i \in I$

$\Leftrightarrow (E \cap Y) \times (F \cap X_{i})$ open in $Y \times X_{i}$, $\forall i \in I$.

Now we have $E \cap Y = E$, which is open in $Y$, and $F \cap X_{i}$ is open in $X_{i}$.

So $h^{-1}(E \times F)$ is open in $\coprod (Y \times X_{i}).$

We now prove that $h$ is open. Let $A \subseteq \coprod (Y \times X_{i})$ open. This means that $A \cap (Y \times X_{i})$ is open in $Y \times X{i}$, for every $i \in I$. $A$ is a basis open subset if it is of the form $E \times F$ with $E$ open in $Y$ and $F$ open in $X_{i}$ for every $i$ in $I$. Now we have

$h(A)$ open in $Y \times \coprod X_{i}$ $\Leftrightarrow E \times F$ open in $Y \times \coprod X_{i} \Leftrightarrow$ $E$ open in $Y$ and $F$ open in $X_{i}$, for all $i \in I$.

Hence, $h$ is a homeomorphism.

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There are continuous bijections that are not homeomorphisms. –  Hurkyl Oct 29 '12 at 19:05
I thought it would be clear that this bijection is open... I edited the proof and showed that it is indeed open, hence a homeomorphism. –  KarenVO Nov 6 '12 at 12:28