Let $X,Y,X^*,Y^*$ be topological spaces such that $X$ is homeomorphic to $X^*$ and $Y$ is homeomorphic to $Y^*$. Prove that $X\times Y$ is homeomorphic to $X^*\times Y^*$.
I know similar questions have been asked on here before, but I was wondering if the way I proved it is correct because I feel like I "cheated". The fact that the product spaces are bijective is easy enough, I am having doubts about proving continuity! Any help at all is greatly appreciated :)
Here is the part of my proof (Assuming I proved Bijectivity)
Need to show that $f\times g : X \times Y \to X^* \times Y^*$ defined by $(x,y)\to \left(f(x),g(y)\right)$ is Continuous:
I will define the following:
$p_1:X\times Y \to X$ projection map
$p_2:X\times Y \to Y$ projection map
$p_1^*:X^*\times Y^* \to X^*$ projection map
$p_2^*:X^*\times Y^* \to Y^*$ projection map
Then I say that :
\begin{align} (f\times g)(X\times Y)&=p_1^{-1*}\left( f\left(p_1(X\times Y)\right) \right) \cap p_2^{-1*}\left(g\left(p_2(X\times Y)\right)\right) \\ &=\left(p_1^{-1*}\circ f \circ p_1\right)(X\times Y) \cap \left(p_2^{-1*}\circ g \circ p_2\right)(X\times Y) \end{align}
$$AND$$
\begin{align} (f\times g)^{-1}(X^*\times Y^*)&=\left(p_1^{-1*}\circ f \circ p_1\right)^{-1}(X^*\times Y^*) \cap \left(p_2^{-1*}\circ g \circ p_2\right)^{-1}(X^*\times Y^*) \\ &=\left(p_1^{-1}\circ f^{-1} \circ p_1^{*}\right)(X^*\times Y^*) \cap \left(p_2^{-1}\circ g^{-1} \circ p_2^*\right)(X^*\times Y^*) \end{align}
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Let $U^*\in \mathcal T_{X^*}$ and $V^*\in \mathcal T_{Y^*}$, so $U^*\times V^*$ open in product space $X^*\times Y^*$.
\begin{align} (f\times g)^{-1}(U^*\times V^*)&=p_1^{-1}\left(f^{-1}\left(p_1^*(U^*\times V^*)\right)\right)\cap p_2^{-1}\left(g^{-1}\left(p_2^*(U^*\times V^*)\right)\right) \\ &=p_1^{-1}\left(f^{-1}(U^*) \right)\cap p_2^{-1}\left(g^{-1}(V^*)\right) \end{align}
Since $f$ is continuous on $X$, and $U^*$ open in $X^*$, then $f^{-1}(U^*)$ is open in $X$.
Clearly $p_1^{-1}\left(f^{-1}(U^*) \right)=f^{-1}(U^*)\times Y$ is open in $X\times Y$.
Since $g$ is continuous on $Y$, and $V^*$ open in $Y^*$, then $g^{-1}(V^*)$ is open in $Y$.
Clearly $p_2^{-1}\left(g^{-1}(V^*) \right)=X\times g^{-1}(V^*)$ is open in $X\times Y$.
The intersection of two open sets in $X\times Y$ is again open by definition. So $(f\times g)^{-1}(U^*\times V^*)$ is open in $X\times Y$, so $f\times g$ is continuous on $X\times Y$.
Your proof is correct, but you can do it simpler. The product topology $\mathcal{T}_{X \times Y}$ is defined as the coarsest topology on the set $X \times Y$ for which $p_1, p_2$ are continuous. A well-known consequence is that a function $\phi : Z \to X \times Y$ defined on a topological space $Z$ is continuous if and only $p_1 \circ \phi : Z \to X$ and $p_2 \circ \phi : Z \to Y$ are continuous.
Now let $f : X \to X^*$ and $g : Y \to Y^*$ be continuous maps. Then $f \times g$ is continuous because $f \circ p_1$ and $g \circ p_2$ are continuous and $p^*_1 \circ (f \times g) = f \circ p_1$ and $p^*_2 \circ (f \times g) = g \circ p_2$.