$X \times \{y\}$ is homeomorphic to $X$ Can I get a proof verification? The only thing I am unsure about is the proof of continuity of $f,f^{-1}$. I know the proof of bijectiveness is trivial.
Prove:$X \times \{y\}$ is homeomorphic to $X$.
Attempt: Take the fact that we can prove continuity by showing that the inverse image of basis elements are open. Define the map $f:X \times \{y\} \rightarrow X$ by $f(x,y)=x$. Then $f$ is one to one since $f(x,y)=f(w,y)\implies x=w$ and so $(x,y)=(w,y)$. $f$ is surjective, since for any $x \in X, f(x,y)=x$. So $f$ is bijective. Let $U$ be an open set in $X$. Then $f^{-1}(U)=U \times \{y\}=(X \times \{y\}) \cap (U \times Y)$, which is open in the subspace topology, so $f$ is continuous.Let $W$ be any open set in $X \times \{y\}$. Then $W=(U \times V) \cap (X \times \{y\})$ where $U \times V$ is open in the product topology. Then $f(W)=U$ is open in $X$ and so $f^{-1}$ is continuous.

I know you didn't ask for it, but I've nothing better to do, and have been thinking about it.  The open sets of $X$ and $X\times\{y\}$ are in bijective correspondence, I think rather trivially.   The underlying sets are also clearly in bijective correspondence. Thus they are homeomorphic.  That's my alternative proof.
