Is $\mathrm{Aut}$ a functor/invariant?

Question

Can the following assignment on objects be made into a functor from the category of topological spaces into the category of groups? Each topological space $X$ gets mapped to $\mathrm{Aut}(X)$, its automorphism group (group of all homeomorphisms from $X$ to $X$). I know that one necessary condition of functors $F\colon \mathcal C\to\mathcal D$ is that whenever $X, Y\in\mathcal C$ are isomorphic, then $F(X)$ and $F(Y)$ are isomorphic. This should be true for $F = \mathrm{Aut}$.

I read that in algebraic topology one is interested in so-called algebraic "invariants". Is $\mathrm{Aut}$ such an invariant (after all, it assigns to each topological space a group)?

Answer 1

First off, it is true that the automorphism group is an invariant in the sense that if two spaces $X$, $Y$ are homeomorphic, then their automorphism groups are isomorphic.

To see this, for a homeomorphism $f: X \rightarrow Y$ consider $\Phi(f): \text{Aut}(X) \rightarrow \text{Aut}(Y)$ via $\varphi \mapsto f \circ \varphi \circ f^{-1}$.

However, in order to make the assignment $F: X \mapsto \text{Aut}(X)$ into a functor you of course need to specify what $F$ does on morphisms $f: X \rightarrow Y$. The naive choices would be pre- or post- composition, but neither of those work since that would not give back an automorphism, but a continuous map between the spaces.

So, maybe there is a way to make it into a functor, but the obvious candidates do not work.

Edit regarding comment: the functor that Kevin Arlin is referring to is defined as follows: Let $\text{Core}(\mathbf{Top})$ be the category with objects topological spaces and maps the homeomoprhisms. Then define the functor $\Phi: \text{Core}(\mathbf{Top}) \rightarrow \mathbf{Grp}$ on objects as $\Phi(X) = \text{Aut}(X)$ and on morphisms $f: X \rightarrow Y$ as $\Phi(f): \text{Aut}(X) \rightarrow \text{Aut}(Y)$ via $\varphi \mapsto f \circ \varphi \circ f^{-1}$.