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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?

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  • $\begingroup$ 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 $X^+$. $\endgroup$ – freakish Dec 13 '18 at 16:15
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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 $X' = (-1,1), Y' = S^1 \cup [1,2) \times \{ 0 \}$.

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