Of course, Peter is correct below, since Gi is not part of the diagram (although Fi is). One more thing. The reason that B^n --> B^{n+1} is monic (which is crucial) is that they are subgroups of F^n and F^{n+1} resp. One thing I found curious is that in any commutative diagram with exact rows 0 ----> A -----> B -----> C -----> 0 | | | | | | | | | v v v 0 ----> A' ----> B' ----> C' ----> 0 if C ---> C ' is monic, then so is A' +_A B ---> B'. Well it turns out that there is an exact sequence 0 ---> ker (C --> C') ---> A' +_A B ---> B' I don't see this as coming from the snake lemma, but who knows. ---------- Forwarded message ---------- Date: Tue, 4 Apr 2000 09:33:34 -0400 (EDT) From: Peter Freyd <pjf@saul.cis.upenn.edu> To: barr@barrs.org Subject: just to be sure The first inequality below should be j < i? The arrows in the diagram are all the alpha j, for j =< i, as well as all the arrows from the restrictions of F and G. The diagram has a cocone to Gi and that cocone is monic (meaning the natural map from the colimit to Gi is monic) for each i in I, call alpha supernatural. Now the condition is that whenever alpha is supernatural, the induced colim F --> colim G is monic.