* just to be sure (fwd)
@ 2000-04-04 15:08 Michael Barr
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From: Michael Barr @ 2000-04-04 15:08 UTC (permalink / raw)
To: Categories list
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.
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2000-04-04 15:08 just to be sure (fwd) Michael Barr
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