Subject: categories: just to be sure (fwd)

From: Michael Barr <[email-masked]>
Date: Tue, 4 Apr 2000 11:08:32 -0400 (EDT)

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 <[email-masked]>
To: [email-masked]
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.