Homeomorphism between $Y$ and $\{b\}\times Y$

Question

If I have two topological spaces $(X,\mathcal{T}_x)$ and $(Y,\mathcal{T}_y)$. I am trying to show that the space $\{b\}\times Y$ is homeomorphic to Y, where $b\in X$

It is my understanding that if I find a function between the two spaces:

$$T: \{b\}\times Y\longrightarrow Y$$

that is continuous, bijective and $T^{-1}$ is also continuous then the two spaces are homeomorphic.

I figured that

$$T(b, y)=y$$

would suffice. I am self teaching myself topology. Is my function enough to justify that the two topological spaces are homeomorphic?

Answer 1

Yes, your $T$ is the restriction (to $\{b\} \times X_2$) of the second projection $\pi_2: X_1 \times X_2 \to X_2$, which is continuous by the definition of the product topology, and restrictions of continuous functions are still continuous.

Its inverse is $j: X_2 \to X_1 \times X_2$ defined by $j(x)=(b,x) \in \{b\} \times X_2$, which is continuous as $\pi_1 \circ j$ is constantly $b$ (so continuous) and $\pi_2 \circ j$ is the identity on $X_2$ (so continuous too), using the universal property of continuity of $X_1 \times X_2$. So $T$ is continuous with continuous inverse and so a homeomorphism.

Answer 2

The open sets in $X \times Y$ are by definition unions of products $U \times V$ with $U$ open in $X$ and $V$ open in $Y$.

The open sets in $\{x\}\times Y$ are intersections of $\{x\}\times Y$ with these open sets. That is they are unions of sets of the form: $$ (\{x\}\times Y)\cap (U \times V)$$

Such a set will either be: $$(\{x\}\times V)$$ or empty depending on whether or not $x\in U$. Such sets are closed under union so any open set in $X \times Y$ has the form $(\{x\}\times V)$, for some open set $V$ in $Y$.

Conversely all such sets are open in $\{x\}\times Y$: $$ (\{x\}\times V)=(\{x\}\times Y)\cap (X \times V)$$

Thus under your identification the open sets of $Y$ and $\{x\}\times Y$ are the same.

Answer 3

Indeed, $\{$x$\}$ $\times Y\approx Y$, where $x\in X$ and $X$ is a space. Note that $X\times Y$ is assumed to be equipped with the product topology.

The map $T: \{x \} \times Y\rightarrow Y$ given by $T(x,y)=y$ is indeed a homeomorphism.

Let $\pi: X\times Y\rightarrow Y$ denote the projection map. Clearly, $\pi$ is $T$ when the domain is restricted to $ \{x \} \times Y$. As a consequence, continuity follows. The inverse of $T$ is $T^{-1}(y)=(x,y)$ . Since $\pi \circ T^{-1}$ (this is the restricted projection map) is the identity map on $Y$, it follows that $T^{-1}$ is continuous. Therefore we conclude that the two spaces are homeomorphic via the map $T$.