A number of topological invariants take the form of functors $\mathscr{T}\to\mathscr{G}$, where $\mathscr{T}$ is the category of all topological spaces and continuous functions, and $\mathscr{G}$ is the category of all groups and homomorphisms. For examples, consider the homology groups $H_{n}(X)$ or the homotopy groups $\pi_{n}(X)$. Of course, a problem with these invariants is that they are not fully faithful functors, i.e., $H_{n}(X)\cong H_{n}(Y)$ does not imply that $X$ and $Y$ are homeomorphic. The existence of a fully faithful functor $F:\mathscr{T}\to\mathscr{G}$ would imply that $\mathscr{G}$ has a subcategory $F\mathscr{T}$ equivalent to $\mathscr{T}$. This would be both rather disturbing and extremely interesting. First, it would mean that in a sense, all of topology is just a subset of group theory, which would be rather disturbing to topologists, but it would also reveal a fundamental connection between two seemingly disparate disciplines. My question is: is it possible to prove that no such functor exists? In other words, could one exhibit some categorical property that $\mathscr{T}$ posesses that $\mathscr{G}$ does not. This question can naturally be extended to other important categories, like $\mathscr{M}$, the category of all modules, or $\mathscr{R}$, the category of all rings. So in general, given arbitrary categories $\mathscr{C}$, $\mathscr{D}$, is there any natural way of showing that no fully faithful functor $F:\mathscr{C}\to\mathscr{D}$ exists, i.e., are there any nice "categorical invariants?"

EDIT: A couple people pointed out that I really ought to be discussing fully faithful functors, rather than just faithful functors. Also, I have changed the title in accordance with Martin Brandenburg's recommendation.

forgeta lot of structure is a fundamentally useful thing, it's in a sense what we aim for since it allows actual reduction of problems to tractable problems. – Ryan Budney Sep 30 '10 at 1:42