# Showing that 2 topological spaces are homeomorphic

Let $X$ and $Y$ be two topological spaces.

I want to show that $\forall y \in Y$ the subspace $X \times \{y\}$ of $X \times Y$ is homeomorphic to $X$.

Attempt:

In other words, this means, I have to show that there exists a function $f: X \times \{y\} \rightarrow X$ that is a homeomorphism. That is: $f$ is continuous/ $f^{-1}$ is continuous/ f is bijective.

If we consider the function: $f: X \times \{y\} \rightarrow X$ defined by: $(x,y) \mapsto x$

It is obviously bijective but how do I show the continuity of $f$ and $f^{-1}$?

• Nov 11 '16 at 18:59

Defined the map $\phi: X \times \{y\} \to X$ by $\phi(x,y) = x$ i.e $\phi = \pi_1$ which is the projection map on $X \times Y$. Recall that this map is continuous since if $U \subset X$ is open then $\pi^{-1}(U) = U \times Y$ which is open in the product topology on $X \times Y$. Now consider the inverse, given by $g: x \mapsto (x,y)$ which is the inclusion map. Let $V \subset X \times \{y\}$ be open then by the basis for the product topology:

$$V = \underbrace{\bigcup_{\alpha} U_{\alpha}}_{V'} \times \{y\}$$

i.e $g^{-1}(V) = V'$ which is open in $X \Rightarrow \phi$ is a homeomorphsim.

• Could you specify what $\alpha$ is? Nov 11 '16 at 19:11
• $\alpha$ is just an index in an arbitrary index set $\mathcal{A}$. By basis for product topology, products of unions of open sets in $X$ and $\{y\}$ but the only open sets in $\{y\}$ are itself and the emptyset. Nov 11 '16 at 19:14

Your function $f:X\times \{y\} \to X$ via $(x,y) \mapsto x$ is just a projection. In particular projections are always continuous (think about the definition of the product topology and think about what a pre-image of an open set in $X$ looks like under the pre-image of this projection).

To go the other direction, what about the natural inclusion map $g_y: X \to X\times Y$ via $x \mapsto (x,y)$? Can you show this is continuous similarly?

If $U$ is open in $X$, then $f^{-1}(U) = \{(x,y): x \in U\} = U \times \{y\}$, which is open. Also, if $V$ is non-empty and open in $X \times \{y\}$, then $V = U\times \{y\}$, where $U$ is open in $X$. So $f(V) = U$ is open in $X$

Recall that the open sets in the product topology of $\Pi_{a \in A} X_a$ are those of the form: $\Pi_{a \in B} O_a \times \Pi_{a \notin B} X_a$, where $B$ is a finite subset of $A$ and each $O_a$ is open in $X_a$.