Is one point compactification functorial? In a algebraic topology course, I recently saw the proof of the theorem that the $n-$ and $m-$ dimensional vector spaces, $n\neq m$, are not isomorphic (in $\mathbf{Top}$). The proof used the compactification of both of these spaces (which are $n-$ and $m-$spheres respectively) and then the reduced homology functor which shows that these are not isomorphic. For this to work it is necessary that the compactification preserves isomorphisms. It was shortly (in elementary terms) explained why that is the case. I wonder now wether the compactification is functorial in general, or wether it just preserves isomorphisms in this specific case.
tl;dr: Question is the title.
 A: As Henno Brandsma said in a comment, it is not functorial.
The Alexandroff one-point compactification of an arbitrary space $X$ is defined as the set $X^* = X \cup \{\infty\}$ with a point $\infty \notin X$. The topology on $X^* $ consists of all open subsets of $X$ and all sets of the form $X^* \setminus C$, where $C \subset X$ is closed and compact. The sets $X^* \setminus C$ are precisely the open  neigborhoods of $\infty$.
Let us consider a map $f : X \to Y$. When does it extend to a continuous $f^*  : X^* \to Y^*$ such that $f^*(\infty) = \infty$?
Clearly $f^*$ is continuous iff for each closed compact $D \subset Y$ we have $(f^*)^{-1}(Y^* \setminus D) = X^* \setminus C$ for some closed compact $C\subset X$. But $(f^*)^{-1}(Y^* \setminus D) = X^* \setminus f^{-1}(D)$, thus $f^*$ is continuous iff preimages of closed compact $D \subset Y$ under $f$ are compact (they are automatically closed by continuity). Such maps $f$ are called proper. Therefore the one-point compactification is a functor from the category of topological spaces and proper maps to the category of compact spaces and continuous  maps.
Note that all homeomorphisms are proper. In contrast, constant maps are never proper unless the domain $X$ is compact.
