# Thread: Is "is homomorphic to" an equivalence relation?

1. ## Is "is homomorphic to" an equivalence relation?

Is "is homomorphic to" an equivalence relation? (Hint: the difficulty is to decide on an appropriate meaning for the quoted phrase.)

Either yes (trivially) or no (nearly trivially).
If "is homomorphic to" is taken to mean there is a homomorphism from into (but not necessarily onto) , then every space is homomorphic to every other space as a zero map always exists.

If "is homomorphic to" is taken to mean there is an onto homomorphism from to then the relation is not an equivalence. For instance, there is an onto homomorphism from to (projection is one) but no homomorphism from onto , so the relation is not reflexive.

Can you help with showing where (in both cases) are the three properties of being symmetric, reflexive and transitive?
Thanking you in anticipation

2. ## Re: Is "is homomorphic to" an equivalence relation?

it depends on "what a homomorphism is". since you appear to be talking about vector spaces, in that context a homomorphism is a linear mapping. usually "homomorphic to" in this context means:

there is a linear map L:V-->W with im(L) = W.

this is clearly reflexive: V~V since we can take L(v) = v.

it is not symmetric: we always have V~0, since we can take L(v) = 0. however, if dim(V) > 0 there is NO linear map L:{0}-->V, since no matter how we define L, im(L) = {0}, so there are no basis vectors at all in im(L), and V certainly has some (since dim(V) > 0).

it is transitive: suppose U~V, and V~W. so we have linear maps L:U-->V and M:V-->W that are onto. given any w in W, there is some v in V with M(v) = w, since M is onto. given any v in M (including the one (or more) with M(v) = w), we have some u in U with L(u) = v.

consider the map M∘L:U-->W. for any w in W, pick u so that L(u) = v, and M(v) = w (as above).

then M∘L(u) = M(L(u)) = M(v) = w. this makes sense because im(L) = dom(M).

if "homomorphic to" merely means a homomorphism L:V-->W exists, then:

V~V (we can still use L(v) = v).

V~W=> W~V (we can always use the 0-map from W to V).

U~V & V~W => U~W (we can always use the 0-map from U to W).

what makes the first definition trivial is that V~W for ANY choice of V and W. all it really says is "any vector space V has a 0-subspace, {0}", which does not allows us to distinguish between vector spaces at all (the equivalence class of V is EVERY vector space, over ANY field).

a little thought shows that to keep "homomorphic to" as some NON-trivial equivalence, we need to fix symmetry, that is: we need to have an "inverse homomorphism". what do we call onto homomorphisms that have inverses?

3. ## Re: Is "is homomorphic to" an equivalence relation?

I'm grateful to you for your help!!! If you allow me, I will ask a few questions:

1. May it be not necessarily for linear map to be reflexive as it appears on the answer- when
there is an onto homomorphism from R3 to R2 (projection is one) but no homomorphism from R2 onto R3, so the relation is not reflexive.
?

2.
V~W=> W~V (we can always use the 0-map from W to V).
Does it mean that- pick any map from V to W as L(v)=w and then use the zero map to come back to the departure space (V), as it will (like any other) always contain the zero? And, if so, is L the zero map, otherwise it causes to skip between relations?

3.
what do we call onto homomorphisms that have inverses?
bijective

4. ## Re: Is "is homomorphic to" an equivalence relation?

1. first of all it's not the maps which have the properties of reflexive, symmetric, transitive, it's the relation ~ defined by "is homomorphic to".

reflexive, for a relation ~ (on some set S), means: for all a in S, a~a. you are confusing this with SYMMETRY: for all a,b such that a~b, b~a. the answer you originally posted used reflexive incorrectly.

2. no L need not be the 0-map (but it might be). symmetry just says: IF V~W, then W~V. symmetry (of ~) says nothing about V,W if they aren't related at all.

3. not quite the right answer. onto homomorphisms that have inverses are bijective, but the fact that they are homomorphisms as well make them something even more special: isomorphisms.

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