Homomorphism as a Structure-preserving Map.

SteveL27

If anything's an analog with homeomorphism, it would be isomorphism, not homomorphism. Homeomorphism is a bijection, as is an isomorphism. There's no topological analog for homomorphism.

Here's something you can do with homeomorphisms that you can't do with isomorphisms or homomorphisms.

X and Y are topological spaces. X is homeomorphic to a proper subset of Y. Y is homeomorphic to a proper subset of X. Must X and Y be homeomorphic?

This seems like a tricky problem until you realize the answer is totally obvious.

I don't believe you can do an analogous trick with algebraic isomorphisms or homomorphisms.

jgens

The answer is no in both cases:
In the topological case take X = R with the usual topology and Y = [0,1] with the subspace topology. Then X is homeomorphic to (0,1) ≤ Y and Y is homeomorphic to [0,1] ≤ X.

In the algebraic case take G = F₂ and H = F₃; that is, G is the free group on 2 elements and H is the free group on 3 elements. Then G is obviously isomorphic to a proper subset of H and it turns out (surprisingly) that H is isomorphic to a proper subset of G.

Edit: You use a different 'trick' in each case, but that is unsurprising since we are considering different kinds of structures. The fact is that you actually do get a similar result with isomorphisms in place of homeomorphisms. There are plenty of cases where you can results that hold for group isomorphisms but not for homeomorphisms (and vice versa), but this isn't one of them.

Number Nine

We're comparing continuous maps with homomorphisms here. Bijective and invertible continuous maps (homeomorphisms) would be analogous to bijective (and symmetric) homomorphisms (isomorphisms), agreed.

SteveL27

Good one, I never realized you could do that algebraically.

(edit) Can you give a hint as to how we can see the iso between F₃ and a proper subset of F₂?

Fredrik

OK, I see now that I didn't think it through. I thought that we would be able to apply the usual ideas about what a structure-preserving map between two structures of the same type is, since a topological space consists of a set and a collection of subsets (=unary relations*) of that set. Hm, I think maybe I understand the problem. It might make sense to consider topological spaces "structures", but since they can involve arbitrarily many unary relations, we can't consider two arbitrary topological spaces to be "of the same type". They don't always have the same number of n-ary relations and n-ary operations, as two structures must in order to be considered "the same type".

*) One way to define an n-ary relation over a set X is as a subset of X × ... x X (n copies of X). So a subset of X is a 1-ary (unary) relation over X.

I suppose that maybe we could make it work if we restricted our attention to topological spaces whose topologies have the same cardinalities. But then we wouldn't end up with one category of topological spaces, but one category of topological spaces for each cardinal number. This seems less useful than the standard definition of the category of topological spaces.

Maybe this is the sort of thing that made mathematicians prefer to talk about categories instead of structures.


Edit: Even if we're only dealing with topological spaces whose topologies have the same cardinalities, open maps wouldn't (always) be structure preserving. For example, if the topologies are infinite but countable, then a structure-preserving map ##f:X\to Y## would be a map that for each positive integer n takes the nth open subset of X to the nth open subset of Y.

Hurkyl

There's a trick that works with any category C: the "structure" of an object X is the functor Hom(-,X). Or equivalently, it is the right C-class (a class with a right action of C on it!) of all morphisms of C whose codomain is X.

By the Yoneda lemma, a morphism in C from X to Y is the "same" thing as a natural transformation Hom(-,X) to Hom(-,Y), which is the "same" thing as a homomorphism of the corresponding C-classes. So the morphisms from X to Y are precisely the structure-preserving maps.

IIRC, in many categories this simplifies. e.g. if C is the category of groups, I think all you need are the morphisms from the free group on 1 element and from the free group on two elements. For the category of manifolds, I think you can manage with just the continuous maps from Euclidean spaces.