The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Epimorphism Proof

Date: 1/12/98 at 17:10:15
From: Doug Haessig
Subject: Category Theory

What is the proof (or at least the idea) behind showing, in the 
category of groups, that an epimorphism is just an onto homomorphism?

Doug Haessig

Date: 1/12/98 at 12:05:13
From: Dr. Joe
Subject: Re: Category Theory

Dear Doug,

The proof is tricky, as is always the case for all such problems. For 
instance, that "an epimorphism between objects from the category of 
Sets(Small) is merely the usual onto map" is quite a tricky thing to 

The proof that I have figured out is a bit long; to cut it short I 
shall prove the harder implication. So please bear with it.


Let f:G --> K be an epimorphism in the categorical sense for two 
groups G and K. Let L = Im f (or actually f(G)). Clearly, elementary 
group theory tells us that L is a subgroup of K.

Now, suppose that f is not surjective. Then, there exists an a in K 
but not in L, denoted by a in K\L.

Next, form a set X = {kL | k in K}U{a}. Of course, kL just means the 
left coset obtained by multiplying k on the left of all elements in L.  
(I assume that you know more group theory than I do.)

Then, we define an element sigma from the set S(X), which is the 
permutation group on X, where

               sigma: X --> X

is defined by

              sigma(L) = a
              sigma(a) = L
             sigma(kL) = kL for any other k not in L.

Next, define two parallel morphims p, q:K --> S(X) as follows:

               p(k)(s) = kk'L if s = k'L for some k' in K;
                       = a    if s = a

               q(k)(s) = sigma o p(k) o sigma^(-1)

where o denotes the usual composition and sigma^(-1) is the inverse 
permutation of sigma.

Now, we shall prove a small lemma:

                L = {k in K | p(k) = q(k)}

Proof of lemma:

We show that L is a subset of {k in K | p(k) = q(k)}.
Pick any l from L.
We need to show that p(l) = q(l).

p(l)(s) = lk'L if s = k'L for some k' in K;
        = a    if s = a

On the other hand,

q(l)(k'L) = sigma o p(l) o sigma^(-1)(k'L)

               sigma o p(l) (k'L) if k' in K\L
          = or
               sigma o p(l) (a)   if k' in L.

               lk'L if k' in K\L
          = or 
               L    if k' in L.

          = lk'L

  q(l)(a) = sigma o p(l) o sigma^(-1) (a)
          = sigma o p(l)(L)
          = sigma (L)
          = a

It is clear that p(l) = q(l).

Now we shall show that if k is not in L, then p(k) is not equal to 

But it is true that if k is not in L, then kL is not equal to L.

This would imply that p(k)(L) = kL
                 and  q(k)(L) = sigma o p(k) o sigma^(-1) (L)
                              = sigma o p(k) (a)
                              = sigma (a)
                              = L

Thus, p(k) is not equal to q(k).

So, we are done with the proof of the lemma.

Now, we move on to the main part of the theorem.

Consider the following composition of arrows:
                    f      ----->
                G -----> K        S(X)

in the category of Groups.

For any g in G, f(g) is in L = Im f.

By the lemma above, pf(g) = qf(g) for all g.  Thus, pf = qf.

Since f is epic, p = q.

But, a in K\L and yet p(a) = q(a).  This contradicts the second part 
of the lemma above. Thus, we are led to the conclusion that such an a 
in K\L does not exist. Therefore, K = L = Im f. Whence, f is a 
surjective homomorphism. (proven)

Cheers and good luck.

-Doctor Joe,  The Math Forum
 Check out our web site!   
Associated Topics:
College Modern Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.