Homomorphism as a Structure-preserving Map.

[V0ODO0CH1LD]

The definition of a homomorphism is that it must preserve some algebraic structure, so if I transform a vector space using homomorphism between vector spaces (linear map), the result must be a vector space too, correct?
Now, if "v" and "w" are two vectors in a vector space V, than "(v + w)" must be a vector in V as well. And if "f(v)" and "f(w)" are vectors in a vector space V', than "f(v) + f(w)" must also be in V'.
But the definition of a linear map is that "f(v + w) = f(v) + f(w)". Would't it suffice to say that if "f(v)" and "f(w)" are in the resulting vector space, "f(v + w)" must be as well? Why must "f(v + w)" equal "f(v) + f(w)"?
If my question seems stupid in the context of linear algebra just think of it in terms of groups and homomorphisms between groups. That's where my confusion started anyway.

 

[Number Nine]

DELETED
Read below.

 

[V0ODO0CH1LD]

So even though a homomorphism is a map that preserves an algebraic structure, it need not be the only one that does?
Or are you saying that if a map preserves an algebraic structure is it automatically a homomorphism? In which case how can you prove that if a map does not contain the properties of a homomorphism it breaks the algebraic structure?

EDIT: Also how can you prove that if a map has the properties of a homomorphism (mainly, f(v + w) = f(v) + f(w)) than it preserves the algebraic structure? I guess you could use linear algebra or group theory for examples. I am sort of a noob at this, and those are the only two concepts I am familiar with so far.

 

[Number Nine]

So even though a homomorphism is a map that preserves an algebraic structure, it need not be the only one that does?
Or are you saying that if a map preserves an algebraic structure is it automatically a homomorphism? In which case how can you prove that if a map does not contain the properties of a homomorphism it breaks the algebraic structure?

First of all, ignore my previous post; I misread "homomorphism" as "homeomorphism" and as a result everything I said was nonsense. Onward...

Now, if "v" and "w" are two vectors in a vector space V, than "(v + w)" must be a vector in V as well. And if "f(v)" and "f(w)" are vectors in a vector space V', than "f(v) + f(w)" must also be in V'.

Yes. This is necessarily true because vector spaces are closed under addition.

But the definition of a linear map is that "f(v + w) = f(v) + f(w)". Would't it suffice to say that if "f(v)" and "f(w)" are in the resulting vector space, "f(v + w)" must be as well? Why must "f(v + w)" equal "f(v) + f(w)"?

A linear map is really just what we call a homomorphism between vector spaces. Their definitions are thus the same. You can relax the definition and consider functions where f(v + w) does not necessarily equal f(v) + f(w), but these behave very differently. You can see this for yourself by comparing some nicely behaved linear transformations in the plane like rotation and reflection matrices with some arbitrary non-linear functions. In an algebraic sense, the collection of linear transformations on a vector space behave very nicely, which is why we give them a name.

Or are you saying that if a map preserves an algebraic structure is it automatically a homomorphism?

A map that has the property that f(v + w) = f(v) + f(w) is necessarily a homomorphism, so this is a true. There are plenty of mappings that aren't homomorphisms, but these mappings don't preserve structure (not all of it).

In which case how can you prove that if a map does not contain the properties of a homomorphism it breaks the algebraic structure?

This is a tricky question in that algebraic properties are sort of defined by homomorphisms. Any good abstract algebra text should have a few examples/exercises detailing that homomorphisms map the identity to the identity, map subgroups to subgroups, etc. Gilbert's Elements of Modern Algebra is a good one (and very accessible). If you really want to understand how important and deep they are, Gilbert's text has a nice chapter on the homomorphism theorems.

 

[V0ODO0CH1LD]

Thanks!

 

[Number Nine]

As an aside, defining the term "homomorphism" and then defining another term "homeomorphism" which looks the same and refers to pretty much the same concept in a slightly different context is the singularly most stupid accomplishment in all of mathematics.

 

[Hurkyl]

As an aside, defining the term "homomorphism" and then defining another term "homeomorphism" which looks the same and refers to pretty much the same concept in a slightly different context is the singularly most stupid accomplishment in all of mathematics.

A homeomorphism is not merely a homomorphism of topological spaces: it is an isomorphism of topological spaces.

 

[Hurkyl]

A map that has the property that f(v + w) = f(v) + f(w) is necessarily a homomorphism, so this is a true. There are plenty of mappings that aren't homomorphisms, but these mappings don't preserve structure (not all of it).

I want to expand on this.

If we're considering magmas or semigroups, your statement is directly true.

If we're considering monoids, your statement is actually false! 0 is part of the monoid structure, and so we must additionally have f(0) = 0. Consider the monoid M with two elements 0 and 1, with its operation defined by

• 0 + 0 = 0
• 0 + 1 = 1 + 0 = 1 + 1 = 1

Also consider the map from M to itself defined by

f(x) = 1​

Then f has the property f(x+y) = f(x) + f(y). However, f is not a homomorphism!


If we're considering groups, then inverses are part of the structure too, so we additionally require [itex]-f(x) = f(-x)[/itex]. Peculiarly, the group properties mean that your statement is once again true: if [itex]f(x+y) = f(x) + f(y)[/itex] is an identity, then it turns out that [itex]f(0) = 0[/itex] and [itex]f(-x) = -f(x)[/itex] is automatically true as well.



Now, we were talking about (real) vector spaces, and vector spaces have additional structure beyond that of a group: they have scalar multiplication, which must be preserved by a homomorphism: f(rv) = r f(v). Except for special scalar fields (e.g. if we're doing rational vector spaces instead of real vector spaces), this doesn't automatically follow from f(v+w) = f(v) + f(w). There are maps that preserve addition but don't preserve scalar multiplication.

Such maps are horrible, though: I think you need the axiom of choice to write one down. The case of complex vector spaces is easier to find counterexamples for, such as the map from the one dimensional complex vector space to itself given by:

(a + bi) --> (a + b)​

(where a and b are the real and imaginary components of the relevant complex number) This map has the property that f(x+y) = f(x) + f(y), but it is not a complex-linear map: it is not a homomorphism of complex vector spaces.

 

[Bacle2]

A homeomorphism is not merely a homomorphism of topological spaces: it is an isomorphism of topological spaces.

I hope this doesn't take us too far off--sorry if it does -- but I guess the

topological equivalent of a homomorphism would be an embedding, wouldn't it?

Unfortunately we don't have a nice theorem like one of the homomorphism

theorem allowing us to mod out by the kernel of an onto homeomorphism to get an

isomorphic copy of the image-- meaning the only thing preventing an onto

homomorphism from being an iso. is not being 1-1 .

 

[micromass]

I hope this doesn't take us too far off--sorry if it does -- but I guess the

topological equivalent of a homomorphism would be an embedding, wouldn't it?

Unfortunately we don't have a nice theorem like one of the homomorphism

theorem allowing us to mod out by the kernel of an onto homeomorphism to get an

isomorphic copy of the image-- meaning the only thing preventing an onto

homomorphism from being an iso. is not being 1-1 .

The topological equivalent of a homomorphism would just be a continuous map. The topological equivalent of an injective homomorphism should be an embedding. And the topological equivalent of a surjective homomorphism should be a quotient map.

I write "should be" because comparing algebraic properties with topological properties is pretty tricky. For example, an embedding of topological spaces is not just an injective continuous map, like it is with algebraic objects. There are many subtle categorical differences between algebra and topology.

 

[Fredrik]

The topological equivalent of a homomorphism would just be a continuous map.

Shouldn't that be an "open map"? I'd say that the structure to be "preserved" by a function ##f:X\to Y## is the choice of what subsets to call "open". If f(E) is open for all open ##E\subset X##, then f is said to be an open map. If ##f^{-1}(E)## is open for all open ##E\subset Y##, then f is said to be continuous. So a continuous map seems to be "structure preserving" in the wrong direction.

 

[Bacle2]

Right: a continuous map does not preserve any structure that I can tell. Actually, it

is sort of the other way-round: a reasonable similarity of structure must exist for

a non-trivial (non-constant) continuous map between spaces is possible, e.g., If X is

connected and Y is totally-disconnected. But, yes, in my own post, an embedding should

correspond to a 1-1 homomorphism.

 

[Number Nine]

A homeomorphism is not merely a homomorphism of topological spaces: it is an isomorphism of topological spaces.

My statement was an expression of frustration over misreading one as the other, not an attempt at definition.

 

[micromass]

Shouldn't that be an "open map"? I'd say that the structure to be "preserved" by a function ##f:X\to Y## is the choice of what subsets to call "open". If f(E) is open for all open ##E\subset X##, then f is said to be an open map. If ##f^{-1}(E)## is open for all open ##E\subset Y##, then f is said to be continuous. So a continuous map seems to be "structure preserving" in the wrong direction.

Hmmm, that's a good point.

But firstly, I was talking about a categorical interpretation of things. A mathematical structure should always induce a category. A category consists of certain objects (such as groups, vector spaces and topological spaces). And between these objects, we should have the "interesting maps". Now, in algebraic contexts, the interesting maps are indeed structure preserving maps. But in topological contexts, the interesting maps are not the open maps, but rather the continuous maps. Indeed, we invented topology in exactly a way so that we could still make sense of continuous maps.

So, if you want to call homomorphisms the "interesting maps", then the topological homomorphisms should be the continuous maps. But they are indeed not structure preserving.

However, it is curious that open maps are also not structure preserving. For example, the image of a closed set is not a closed set under an open map. Or the image of a compact set is also not a compact set. So I wouldn't immediately call the open maps structure preserving since many of the interesting properties in topology are not preserved.

 

[Hurkyl]

Continuous maps are the maps that preserve the structure of topological spaces.
Okay, so what is the structure of a topological space?
Precisely what is preserved by continuous maps.

While I say it glibly, the meaning is serious: at some point, it becomes less useful to have a prior notion of what the "structure" of something is, and define what the homomorphisms are. And then, the right notion of structure is whatever is derived from that.

 

[Number Nine]

From a categorical point of view, aren't morphisms between topological spaces continuous maps? That would seem to make them the topological analog of a homomorphism (at least from the perspective of category theory).

 

[Bacle2]

Continuous maps are the maps that preserve the structure of topological spaces.
Okay, so what is the structure of a topological space?
Precisely what is preserved by continuous maps.

While I say it glibly, the meaning is serious: at some point, it becomes less useful to have a prior notion of what the "structure" of something is, and define what the homomorphisms are. And then, the right notion of structure is whatever is derived from that.

You're right, but shouldn't the right notion agree with some ideas of what the basic

makeup of a topological space is? Most (point set, at least) topological properties are

defined in terms of open sets: compactness,connectedness, notions of convergence,etc.

Moreover, these properties are transferred between spaces with comparable topologies.

So it seems reasonable to define the open set as at least one of the basic components* --

if not _the_ main component of the basic makeup of a topological space.

*Maybe a bad choice; obviously used in its everyday meaning, not the topological sense.

 

[SteveL27]

From a categorical point of view, aren't morphisms between topological spaces continuous maps? That would seem to make them the topological analog of a homomorphism (at least from the perspective of category theory).

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]

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?

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

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]

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.

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]

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.

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]

A mathematical structure should always induce a category.

However, it is curious that open maps are also not structure preserving. For example, the image of a closed set is not a closed set under an open map. Or the image of a compact set is also not a compact set. So I wouldn't immediately call the open maps structure preserving since many of the interesting properties in topology are not preserved.

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.