# Locally compact hausdorff space homeomorphism

Suppose that $$X, Y$$ are locally compact Hausdorff space. A bijective $$f:X\to Y$$ is also a homeomorphism?

Let $$X^+=X\cup \{\infty_x\}$$ and $$Y^+=Y\cup\{\infty_y\}$$. By one point compactification, $$X^+$$ and $$Y^+$$ are compact. Define function $$f^*:X^+\to Y^+$$ such that $$f^*|_X=f$$ and $$f^*(\infty_x)=f^*(\infty_y)$$.

Then, $$f^*:X^+\to Y^+$$ is well defined. Since $$f$$ is bijective function, $$f^*$$ is bijective function.

For open subset $$U$$ not containing $$\infty_y$$ in $$Y^+$$, obviously $$f^{-1}(U)$$ is open in Y. Then, $$f^{-1}(U)$$ is open in $$Y^+$$. For open subset $$U$$ containing $$\infty_y$$ in $$Y^+$$, $$f^{-1}(Y^+\setminus U)$$ is closed in X, because $$Y^+ \setminus U$$ is closed and $$f$$ is continuous.

$$f^*$$ is homeomorphic if $$f^*$$ is continuous.

How can I show that $$f^*$$ is continuous and $$X$$ and $$Y$$ are homeomorphic?

• You are wrong that $f^*$ is continuous. For example if $X=[0,1)$ and $Y=[0,1]$ then $X^+=[0,1]$ and $Y^+=[0,1]\cup\{\infty\}$. Such $f^*$ cannot be continuous because $X^*$ is connected while $Y^*$ is not. The particular mistake is that you claim that $f^{-1}(Y^+\backslash U)$ is closed. It is closed in $X$ but not necessarily in $X^+$. The same is for $f^{-1}(U)$. It is closed in $X$, not necessarily in $X^+$. – freakish Dec 13 '18 at 16:15

It is not true. Let $$X = [0,2\pi), Y = S^1$$. Then $$f : X \to Y, f(t) = e^{it}$$, is a continuous bijection, but not a homeomorphism.
You may suspect that the reason for this phenomenon is that $$Y$$ is compact and $$X$$ is not compact. But you can easily modify the above example to $$X' = (-1,1), Y' = S^1 \cup [1,2) \times \{ 0 \}$$.