Epimorphism ProofDate: 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? Sincerely, 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 prove. 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. Proof: 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 q(k). 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: p f -----> G -----> K S(X) -----> q 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! http://mathforum.org/dr.math/ |
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