How to show the one point compactification of a space is unique up to homeomorphism? 
Suppose $X$ is a space with a one point compactification $f: X\to Y$, where $x_0$ is the point that gets added to $X$, and suppose we have another one point compactification $f': X\to Y'$ where $x_1$ is the point that gets added to $X$.

I think we can define a homeomorphism $g: Y\to Y'$ in the following way: $g(x_0)=x_1, g(x)=f'f^{-1}(x)$ for all other $x$.  Intuitively I'm thinking this should be a homeomorphism (if it's not correct me), but I'm not seeing how to prove it.  I think it's fairly clear that it should be bijective, but how to show continuity (of both it and its inverse) isn't coming to me.  If anyone could give a hint that would be swell.
 A: The homeomorphisms $f$ and $f'$ are continuous and their inverses are continuous; therefore $f'f^{-1}$ is continuous and its inverse $ff'^{-1}$ is continuous.
That is enough to show that $f'f^{-1}$ and $ff'^{-1}$ are continuous at every point except $x_0$ and $x_1.$
Continuity of a function $f$ at a point $x$ in its domain can be characterized by saying that for every open neighborhood $V$ of $f(x)$ there is some open neighborhood $U$ of $x$ such that the image of $U$ under $f$ is a subset of $V.$
To show that $f'f^{-1}$ is continuous at $x_0,$ suppose $V$ is some open neighborhood of $f(x_0) = x_1.$ An open neighborhood of the point $x_1$ that gets added is the complement (within $X\cup\{x_1\}$) of a closed set in $X.$ So $X\smallsetminus V$ is closed in $X;$ hence the complement of $X\smallsetminus V$ within $X\cup\{x_0\}$ is an open neighborhood of $x_0.$ It can readily be shown that the image of that open neighborhood of $x_0$ is that open neighborhood of $x_1.$
Continuity of the inverse $ff'^{-}$ is shown in the same way.
