The subspace $\{x\}×Y\subseteq{ X\times Y}$

Question

$\forall x\in X$ consider the subspace $\{x\}×Y\subseteq{ X\times Y}$. Is this subspace homeomorphic to $Y$? Is that true? What is the justification?

Answer 1

Fix $x$

Define $f: Y \to X \times Y$ by $f(y)=(x,y)$. This is continuous, as $\pi_X \circ f$ (a constant map with value $x$)and $\pi_Y \circ f$ (the identity on $Y$) are both continuous, by the universal property for products. Or use that $f^{-1}[U \times V]$ is either empty (if $x \notin U$) or $V$ (otherwise) and thus open for any basic open $U \times V$ in $X \times Y$.

$f$ is obviously a bijection with $\{x\} \times Y$ as its image and its continuous inverse is just $\pi_Y$, restricted to that set.

So $f$ shows $\{x\} \times Y \simeq Y$ for any $x \in X$.

Answer 2

It seems pretty obvious that we can map $Y$ to $\{x\}\times Y$ via $f(y) = (x,y)$.

Is that a homeomorphism?.... (ducks out to google what "homeomorphism" means) ...

Well, the definition of homeomorphism is:

$f$ is bijective, continuous, and $f^{-1}$ is continuous.

....

a) $f$ is onto. Pf: For any $(x,y) \in \{x\}\times Y$ then $f(y) = (x,y)$

b) $f$ is one to one. Pf: If $f(y) = f(w)$ then $(x,y)=(x,w)$ and $y=w$.

c) $f$ is continuous. Aw, c'mon. Pf: $|y-a|=\sqrt{0^2 + (y-a)^2}=D((x,y)-(x,a))=D(f(y),f(a))$ so for any $\epsilon >0$ then $|y-a|< \epsilon \iff D(f(y),f(w)) < \epsilon$ so both $f$ and $f^{-1}$ are continuous.

It's a homeomorphism.