$(X_1,\tau_1)\times(X_2,\tau_2)\cong (X_2,\tau_2)\times(X_1,\tau_1) $

Question

Let $(X_1,\tau_1)$ and $(X_2,\tau_2)$ be topological spaces. Prove that $(X_1,\tau_1)\times(X_2,\tau_2)\cong (X_2,\tau_2)\times(X_1,\tau_1) $.

This is a weird exercise once I can visualize there is a swap of the axis between the two spaces which keeps the spaces being homeomorphic. But I cannot conceive a proof of the statement.

Question:

How should I prove the statement?

Thanks in advance!

Answer 1

Fact: For any space $X$ and any function $f: X \to (X_1, \tau_1) \times (X_2, \tau_2)$: $f$ is continuous iff $\pi_1 \circ f: X \to (X_1, \tau_1)$ and $\pi_2 \circ f: X \to (X_2, \tau_2)$ are both continuous.

Then the swapping map $s(x,y)=(y,x)$ is clearly a bijection (the same formula defines the inverse as well) and $\pi_1 \circ s= \pi_2$ and $\pi_2 \circ s =\pi_1$ so $s$ is continuous (projections are always continuous in the product topology) and the same reasoning can be held for the inverse. So $s$ and its inverse are continuous, and we are done.

Answer 2

Define a map f(a,b)=(b,a) between the spaces and prove that this map is a homeomorphism with respect to the product topology.