Homeomorphism between a subspace of a product topology and one of the factors of product space This is my first question on SE. I will try to be as clear as possible. I have this question as my homework in my topology course.
We have to prove that $X_1 \times \{x_2\} \sim X_1$, where $X_1 \times \{x_2\}$ is considered as subspace of product topology on $X_1 \times X_2$ and $\{x_2\}$ is any arbitrary point in $X_2$.
I can solve it using commutative triangle for product topologies but we are required to use the following results to prove this proposition:
1) $\{x_1\} \times X_2 \sim X_2$ where, $\{x_1\} \times X_2$ is a subspace of $X_1 \times X_2$ and $\{x_1\}$ is any arbitrary point in $X_1$. 
2) $X_1 \times X_2 \sim X_2 \times X_1$.
I think we have to come up with a series of homeomorphisms such that the composition of all these bijections will be the required homeomorphic function. But I can't think of any such composition.
Any help will be much appreciated.   
 A: The homeomorphism $X_1\times X_2:X_2\times X_1:\langle x_1,x_2\rangle\mapsto\langle x_2,x_1\rangle$ gives you a homeomorphism from $X_1\times\{x_2\}$ to $\{x_2\}\times X_1$, and the first result that you’re to use gives you a homeomorphism from $\{x_2\}\times X_1$ to $X_1$; just compose these homeomorphisms.
A: You can use that a map $F$ from any space $Z$ into $X_1 \times X_2$ is continuous iff $\pi_1 \circ F$ and $\pi_2 \circ F$ are both continuous (where $\pi_i$ is the projection from the product onto $X_i$).
So the swap map $F: X_2 \times X_1 \to X_1 \times X_2$ is continuous as $\pi_1 \circ F=\pi_2$ and $\pi_2 \circ F = \pi_1$ and so are both continuous. For the obvious inverse map we use the same criterion but for $X_2 \times X_1$ and $Z=X_1 \times X_2$ instead. 
The swap map $F$ restricted to $X_1 \times \{x_2\}$ gives you a homeomorphism with $F[X_1 \times \{x_2\}]= \{x_2\} \times X_1$, and then the fact 1. you quote gives you another homeomorphism of the latter set with $X_1$. But IMHO it's nonsense to force a student to use a certain pre-set proof method in this setting, as the natural projection map $\pi_1$ (restricted) is already a homeomorphism from $X_1 \times \{x_2\}$ to $X_1$. Why involve the swapping map at all? Seems pointless.
A: 1) f:X×{a} -> X, (x,a) -> x is a homeorphism.
2) g:X×Y -> Y×X, (x,y) -> (y,x). is a homeorphism.  
No compositions are required.
Communitive diagrams just get in the way.
I see no need nor use for them. 
