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Let $ Y $ be a subspace of both a Hausdorff space $ X _ 1 $ and a Hausdorff space $ X _ 2 $ such that $ \mathop { \rm { cl } } _ { X _ 1 } ( Y ) = X _ 1 $ and $ \mathop { \rm { cl } } _ { X _ 2 } ( Y ) = X _ 2 $. Suppose that the identity function $ i : Y \to Y $ can be extended to a continuous function $ f : X _ 1 \to X _ 2 $ as well as to a continuous function $ g : X _ 2 \to X _ 1 $. Prove that $ X _ 1 $ and $ X _ 2 $ are homeomorphic.

I am thinking to prove this by using the properties of continuous function between two topological spaces. But I get stuck in the part $ f ^ { - 1 } $ to prove that is a continuous function.

Thanks for the help.

2 Answers 2

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Note that $g \circ f$ is a continuous function $X_1 \to X_1$ that equals the identity function when restricted to the dense subset $Y \subset X_1$. It follows that $g \circ f$ is the identity function on $X_1$, but the Hausdorff condition is needed for this to be true.

We prove the following lemma: If $\phi,\psi:A \to B$ are continuous functions whose codomain is a Hausdorff space, then $\phi = \psi$ whenever they agree on a dense subset $ C \subset A$. To prove this, we only need to show these functions agree at any $x \in A \backslash C$. Suppose $\phi(x) \neq \psi(x)$. Because $B$ is Hausdorff, there are disjoint open sets $U,V$ about $\phi(x)$ and $\psi(x)$, respectively. By continuity there exists an open set $W \ni x$ such that $\phi(y) \in U$ and $\psi(y) \in V$ whenever $y \in W$. Furthermore, because $x \in \mathsf{closure}(C)$, there exists some $y \in W \cap C$. But this immediately leads to the contradiction $\phi(y) \in U$ while $\psi(y) = \phi(y) \in V$. We deduce $\phi(x) = \psi(x)$, and by the arbitrary nature of $x$ conclude that $\phi = \psi$.

Now apply the argument of the first paragraph to $f \circ g : X_2 \to X_2$, which is continuous and equals the identity function of $Y$.

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    You do need the Hausedorff condition – 2017-02-23
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Let $ X $ be a completely regular space. If $ Y _ 1 $ and $ Y _ 2 $ are two compactifications of $ X $ satisfying the extension property, then $ Y _ 1 $ and $ Y _ 2 $ are equivalent. Can we use the above idea to prove the above problem?