# Surjectivity of epimorphisms of groups

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### Asmodeus

Mar 3, 2000, 3:00:00 AM3/3/00
to
Would someone please explain to me why epimorphisms of groups are
surjective?

Asmodeus

### Arturo Magidin

Mar 3, 2000, 3:00:00 AM3/3/00
to
In article <38c03...@news.arcor-ip.de>,

Asmodeus <asmo...@arcormail.de> wrote:
>Would someone please explain to me why epimorphisms of groups are
>surjective?

Because every subgroup of a group is an equalizer subgroup.

If you want to invoke powerful theorems, you can take Schreier's
theorem on the existence of free products with one amalgamated
subgroup.

But a much easier proof is found in Carl Linderholm's "A Group
Epimorphism is Surjective", in the American Mathematical Monthly 77,
pp. 176-177:

Let f be an epimoprhism from a group G to a group H, and let A be the
image subgroup. We must show that A=H. To do this, construct two
homomorphisms g and k from H to a group K such that gf=kf (morphisms
being written on the left), and use the resulting equation g=k to show
that A=H.

Let H/A be the set of all right cosets of A. Let A' be something which
is not a right coset of A, and let S be the set H/A\cup {A'}. Let K be
the group of all permutations of S. If h, h_1, and h_2 are in H, and
if Ah_1 and Ah_2 are the same coset, then A(h_1h) and A(h_2h) are the
same coset, so the function from A/H to A/H that sends Ah' to A(h'h)
is well defined. It is easily seen to be bijective, its inverse given
by Ah'--> A(h'h_{-1}). If A' is setn to itself, this defines a
bijection of S, which we write as g(h). The function g from H to K
defined this way is actually a homomorphism (excercise left to the
reader). Lets be the permutation of S that interchanges A and A' and
leaves everything else fixed, and if h=in H, write k(h) for the
composite permutation obtained by conjugating g(h) by s. Then the
function k is the composition of g with an inner automorphism of K, so
it is also a homomorphism from H to K.

If a is an element of A, then g(a) leaves both A and A' fixed, and s
leaves every other element of S fixed, so s and g(a) commute, so
g(a)=k(a). Since g and k agree on the range of f, gf=kf. Since f is an
epimorphism, g=k. So g(h) commutes with s for each element h of
H. Since g(h) leaves A' fixed and s exchanges A and A', it follows
that g(h) leaves A fixed. But g(h) sends A to Ah, so h lies in
A. Therefore, A=H, as desired.

Note that this also shows that an epimorphism of ->finite<- groups is
surjective (in the category of all finite groups), which does not

A Theorem of Peter Neumann shows that in any full subcategory C of the
category of all groups in which (a) every group is solvable, and (b) C
is closed under quotients; all epimorphisms are surjective. This was
extended substantially by Susan McKay. On the other hand, from an
example of B.H. Neumann and other work it is easy to exhibit full
subcategories (in fact, varieties) in which not all epimorphisms are
surjective. For example, in the variety generated by A_5, the
immersion A_4->A_5 is an epimorphism.

======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
mag...@math.berkeley.edu

### Arturo Magidin

Mar 3, 2000, 3:00:00 AM3/3/00
to
In article <89pfls\$c8d\$1...@agate.berkeley.edu>,

Arturo Magidin <mag...@yuban.berkeley.edu> wrote:
>In article <38c03...@news.arcor-ip.de>,
>Asmodeus <asmo...@arcormail.de> wrote:
>>Would someone please explain to me why epimorphisms of groups are
>>surjective?
>
>Because every subgroup of a group is an equalizer subgroup.

Sorry for following up on my own post, but I should note that the
statement that every subgroup of a group is an equalizer subgroup
is substantially stronger than the statement that all epimorphisms are
surjective.

For example, in the variety of all nilpotent groups of class two,
every epimorphism is surjective (which follows from the theorem of
Peter Neumann I mentioned in the previous post), yet not every
subgroup is an equalizer subgroup.

For example, take the group given by

G = <x,y | x^{p^2}=y^{p^2}= [x,y]^{p^2}=[x,y,x]=[x,y,y]=e>

and the subgroup H generated by x^p and y^p. It is easy to see that H
is isomorphic to Z/pZ x Z/pZ. However, any pair of maps from G to a
nilpotent group of class two which agree on H will also agree on
[x,y]^p: say f,g:G->K is such a pair; then (note that in this variety,
the commutator bracket acts as a bilinear map from G^ab x G^ab to G'):

f([x,y]^p) = f([x^p,y]) = [f(x^p),f(y)]
= [g(x^p),f(y)] (since x^p lies in H)
= [g(x)^p,f(y)]
= [g(x),f(y)]^p

and by symmetry this is equal to g([x,y]^p). So H cannot be equal to
an equalizer subgroup of G, even though all epimorphisms are
surjective.

### Richard Carr

Mar 3, 2000, 3:00:00 AM3/3/00
to
On Fri, 3 Mar 2000, Asmodeus wrote:

:Date: Fri, 3 Mar 2000 23:17:34 +0100
:From: Asmodeus <asmo...@arcormail.de>
:Newsgroups: sci.math
:Subject: Surjectivity of epimorphisms of groups
:
:Would someone please explain to me why epimorphisms of groups are
:surjective?

An epimorphism is a homomorphism that is surjective (a monomorphism
injective, isomorphism bijective, automorphism bijective between the
object in question and itself, endomorphism homormorphism from the object
to itself)- at least that's the definitions I've seen. Hence, by
definition it is surjective. What definition are you using?

:
:Asmodeus
:
:
:

### Richard Carr

Mar 3, 2000, 3:00:00 AM3/3/00
to
On 4 Mar 2000, Arturo Magidin wrote:

:Date: 4 Mar 2000 00:36:52 GMT
:From: Arturo Magidin <mag...@bosco.berkeley.edu>
:Newsgroups: sci.math
:Subject: Re: Surjectivity of epimorphisms of groups
:
:In article <Pine.LNX.4.21.00030...@cpw.math.columbia.edu>,
:Richard Carr <ca...@cpw.math.columbia.edu> wrote:

:The categorical one.

I see- I never did category theory (excepting the words "you all know
about category theory, right?" delivered so fast that there wasn't time to
respond).

:
:Let me write all funtions on the left.
:
:A monomorphism is a morphism which is left cancellable. That is,
:f:a->b is a monomorphism iff for every pair of morphisms g,h:c->a,
:if fg=fh (meaning the composites are equal), then g=h.
:
:An epimorphism is a morphism which is right cancellable. That is,
:f:a->b is an epimoprhism iff for every pair of morphisms g,h:b->c, if
:gf=hf, then g=h.
:
:An isomorphism is a morphism which has a two-sided inverse.
:
:In many "natural" categories (such as sets and groups), monomorphism
:is equivalent to injective, epimorphism to surjective, and isomorphism
:to bijective. In others it isn't. For example, in the category of
:rings, not every epimorphism is surjective (consider the map from Z to
:Q).
:

I see. Similarly for any field of fractions F={[r/s]:r,s in R, s not 0},
the map h:R->F, h(r)=[r,1] where R is some integral domain.

:Also one should be careful, because monomorphism+epimorphism does not
:imply isomorphism in a general categorical setting; the converse, of
:course, is always true.

Yes. The same example shows this. Since h is injective, it is a
monomorphism as well as en epimorphism but not an isomorphism, since h has
only an inverse on one side. (Presumably you'd want g:F->R but then
h(g([r,s])) cannot equal [r,s] if s isn't 1.)

:
:In the case of objects with underlying sets, it is easy to show that
:every surjective morphism is an epimorphism, adn every injective
:morphism is an epimorphism. If there is also a well-defined notion of
:"free object on one generator", then it is easy to show that any
:monomorphism is necessarily injective. Usually, showing that
:epimorphisms are surjective is the hardest (and often it is not even
:correct).
:
:As another quick example, in the category of Haussdorf topological
:spaces, a continuous map f:T->V is an epimorphism if and only if the
:image f(T) is dense in V.
:
:======================================================================

:
:

### Arturo Magidin

Mar 4, 2000, 3:00:00 AM3/4/00
to
In article <Pine.LNX.4.21.00030...@cpw.math.columbia.edu>,
Richard Carr <ca...@cpw.math.columbia.edu> wrote:
>On Fri, 3 Mar 2000, Asmodeus wrote:
>
>:Date: Fri, 3 Mar 2000 23:17:34 +0100
>:From: Asmodeus <asmo...@arcormail.de>
>:Newsgroups: sci.math
>:Subject: Surjectivity of epimorphisms of groups
>:
>:Would someone please explain to me why epimorphisms of groups are
>:surjective?
>
>An epimorphism is a homomorphism that is surjective (a monomorphism
>injective, isomorphism bijective, automorphism bijective between the
>object in question and itself, endomorphism homormorphism from the object
>to itself)- at least that's the definitions I've seen. Hence, by
>definition it is surjective. What definition are you using?

The categorical one.

Let me write all funtions on the left.

A monomorphism is a morphism which is left cancellable. That is,
f:a->b is a monomorphism iff for every pair of morphisms g,h:c->a,
if fg=fh (meaning the composites are equal), then g=h.

An epimorphism is a morphism which is right cancellable. That is,
f:a->b is an epimoprhism iff for every pair of morphisms g,h:b->c, if
gf=hf, then g=h.

An isomorphism is a morphism which has a two-sided inverse.

In many "natural" categories (such as sets and groups), monomorphism
is equivalent to injective, epimorphism to surjective, and isomorphism
to bijective. In others it isn't. For example, in the category of
rings, not every epimorphism is surjective (consider the map from Z to
Q).

Also one should be careful, because monomorphism+epimorphism does not

imply isomorphism in a general categorical setting; the converse, of
course, is always true.

In the case of objects with underlying sets, it is easy to show that

### Ed Hook

Mar 4, 2000, 3:00:00 AM3/4/00
to
In article <Pine.LNX.4.21.00030...@cpw.math.columbia.edu>,

Richard Carr <ca...@cpw.math.columbia.edu> writes:

|> An epimorphism is a homomorphism that is surjective (a monomorphism
|> injective, isomorphism bijective, automorphism bijective between the
|> object in question and itself, endomorphism homormorphism from the object
|> to itself)- at least that's the definitions I've seen. Hence, by
|> definition it is surjective. What definition are you using?
|>

Presumably, the category-theoretic definition which is that
a morphism a: A --> B is an _epimorphism_ iff fa = ga implies
that f = g for any morphisms f, g: B --> C.

--
Ed Hook | Copula eam, se non posit
MRJ Technology Solutions, Inc. | acceptera jocularum.
NAS, NASA Ames Research Center | I can barely speak for myself, much
Internet: ho...@nas.nasa.gov | less for my employer

### Charles H. Giffen

Mar 7, 2000, 3:00:00 AM3/7/00
to Arturo Magidin
Arturo Magidin wrote:
>
[snip]

>
> In the case of objects with underlying sets, it is easy to show that
> every surjective morphism is an epimorphism, adn every injective
> morphism is an epimorphism.

Is this a typo? (I don't mean "adn" for "and")

If there is also a well-defined notion of
> "free object on one generator", then it is easy to show that any
> monomorphism is necessarily injective. Usually, showing that
> epimorphisms are surjective is the hardest (and often it is not even
> correct).
>

> As another quick example, in the category of Haussdorf

Hausdorff!

topological
> spaces, a continuous map f:T->V is an epimorphism if and only if the
> image f(T) is dense in V.
>

Or, in the category of all topological spaces and continuous
maps, a map f: X -> Y is a monomorphism iff it is an injection,
and it is an epimorphism iff it is a surjection. Hence, for any
space X other than a singleton or the empty set, we have the
maps

X_discrete -> X -> X_indiscrete

where X_discrete is X with every subset open, and X_indiscrete
is X with only the empty set and X open. Each of the maps
above is both epi and mono, but at least one of them is not
a homeomorphism (isomorphism in the category).

For algebraists, another example: Consider associative rings with
unit. The inclusion of the integers in the rationals is both an
epimorphism and a monomorphism, but clearly not an isomorphism.

--Chuck Giffen

### Arturo Magidin

Mar 7, 2000, 3:00:00 AM3/7/00
to
In article <38C53B8D...@virginia.edu>,

Charles H. Giffen <ch...@virginia.edu> wrote:
>Arturo Magidin wrote:
>>
>[snip]
>>
>> In the case of objects with underlying sets, it is easy to show that
>> every surjective morphism is an epimorphism, adn every injective
>> morphism is an epimorphism.
>
>Is this a typo? (I don't mean "adn" for "and")

Yes, I meant "and every injective morphism is a monomorphism."
Sorry. That's what I get for not proof-reading

>> As another quick example, in the category of Haussdorf
>
>Hausdorff!

Argh. I can never keep straight which is the letter that is supposed
to be doubled. Thanks. I'll write it in my blackboard a couple hundred
times.

[.snip.]

>Or, in the category of all topological spaces and continuous
>maps, a map f: X -> Y is a monomorphism iff it is an injection,
>and it is an epimorphism iff it is a surjection. Hence, for any
>space X other than a singleton or the empty set, we have the
>maps
>
> X_discrete -> X -> X_indiscrete
>
>where X_discrete is X with every subset open, and X_indiscrete
>is X with only the empty set and X open. Each of the maps
>above is both epi and mono, but at least one of them is not
>a homeomorphism (isomorphism in the category).

Yes. There is a nice (oldish) article with a bunch of examples, namely
"The meaning of mono and epi in some familiar categories", W. Burgess,
Canad. Math. Bull. 8 (1965) 759-769.

>For algebraists, another example: Consider associative rings with
>unit. The inclusion of the integers in the rationals is both an
>epimorphism and a monomorphism, but clearly not an isomorphism.

In general, for algebras (in the sense of UA), if A->B is any map we
can factor it into a quotient map followed by an immersion. If the
original map is a nonsurjective epimorphism, then the immersion from
this factorization will be an epimorphism and an injection, hence a
monomorphism, but not an isomorphism.

I believe the name for a category in which epi+mono -> iso is
"balanced."