Let $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ be topological spaces. Show that $X \times Y \simeq Y\times X$

Question

The title describes the question. We need to prove that $X \times Y$ is homeomorphic to $Y \times X$. Here's my proof:

To show that the two spaces are homeomorphic we need to establish a continuous bijection between $X\times Y$ and $Y \times X$ that has a continuous inverse. Let $x \in X$ and $y \in Y$ such that $f(x, y) = (y, x)$. In other words, $f$ is a function that swaps the coordinates of its input. It is clear that $f$ is bijective. We need to show that both $f$ and $f^{-1}$ are continuous. Indeed for $f$ to be continuous it suffices to show that the coordinate functions given by \begin{align*} f_1(x, y) &= y \\ f_2(x, y) &= x \end{align*} are continuous. We will prove $f_1$ to be continuous. A similar argument follows for $f_2$. It is obvious that $f_1$ is continuous, since if we choose $V \subseteq Y$ to be open, then $f_1^{-1}(V) \subseteq U_x \times \{y\} $ for any open $U_x \subseteq X$. Thus $f_1$ is continuous. Similarly, one can prove that $f_2$ is also continuous. Now we need to show, $f_1^{-1}$ is also continuous. Choose any open set $U_x \subseteq X$ and $U_y \subseteq Y$. Now $f_1(x, y) = y \in U_y$ for every $x \in U_x$ and $y \in U_y$. Thus $f_1^{-1}$ is point-wise continuous for every $x, y \in U_x \times U_y$. So it is continuous on overall $X \times Y$.

Criticize my proof...

Answer 1

You asked for feedback on your proof, so here goes...

To show that the two spaces are homeomorphic we need to establish a continuous bijection between $X \times Y$ and $Y \times X$ that has a continuous inverse.

Correct.

Let $x \in X$ and $y \in Y$ such that $f (x, y) = (y, x)$. In other words, $f$ is a function that swaps the coordinates of its input. It is clear that $f$ is bijective.

Your idea here is clear (from the second sentence), but the first part is confusing. You haven't defined $f$. You probably mean something like, "consider $f : X \times Y \mapsto Y \times X$ such that $f (x \times y) = y \times x$ for all $x \in X$ and $y \in Y$". Otherwise, it could be interpreted like you are picking a specific $x \in X$ and $y \in Y$ such that $f (x \times y) = y \times x$.

We need to show that both $f$ and $f^{-1}$ are continuous. Indeed for $f$ to be continuous it suffices to show that the coordinate functions given by $f_1 (x, y) = x$ and $f_2 (x, y) = y$ are continuous.

This is correct, but for clarity you should explicitly call out the theorem this implicitly relies on, which is that $f : A \mapsto Y \times X$ defined by $f (a) = f_1 (a) \times f_2 (a)$ is continuous if $f_1 : A \mapsto Y$ and $f_2 : A \mapsto X$ are continuous. In this problem, $A = X \times Y$.

We will prove $f_1$ to be continuous. A similar argument follows for $f_2$. It is obvious that $f_1$ is continuous, since if we choose $V \subseteq Y$ to be open, then $f_1^{−1} (V) \subseteq U_x \times \{ y \}$ for any open $U_x \subseteq X$. Thus $f_1$ is continuous.

This part is not totally clear. What is $y$ and how does it relate to $V$? You can simplify it as follows. Let's start as you did by picking an arbitrary open set $V_Y \subseteq Y$. The preimage of $V_Y$ under $f_1$, as you can check, is $X \times V_Y$. And $X \times V_Y$ is open in $X \times Y$ because (1) $X$ is trivially open in $X$, and (2) $V_Y$ is open in $Y$ by hypothesis. Therefore, $f_1$ is continuous.

Now we need to show, $f_1^{−1}$ is also continuous. Choose any open set $U_x \subseteq X$ and $U_y \subseteq Y$. Now $f_1 (x, y) = y \in U_y$ for every $x \in U_x$ and $y \in U_y$. Thus $f_1^{−1}$ is point-wise continuous for every $x, y \in U_x \times U_y$. So it is continuous on overall $X \times Y$.

This part is also a little confusing. It is not immediately obvious how $f_1^{-1}$ (and $f_2^{-1}$) being continuous implies that $f^{-1}$ is continuous. There is an easier way. The proof so far is enough to show that $f : X \times Y \mapsto Y \times X$ is continuous for arbitrary spaces $X$ and $Y$. Consider $f^{-1} : Y \times X \mapsto X \times Y$ defined by $f^{-1} (y \times x) = x \times y$. It is easy to see that $f^{-1}$ is indeed the inverse of $f$. Note that $f^{-1}$ is continuous by exactly the same arguments as what we made for $f$, since they are defined in the same fashion (just switching the positions of $X$ and $Y$, loosely speaking). So, $f$ is a homeomorphism.