$\coprod_{i \in I} (Y \times X_{i})$ 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!
 A: 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.
