Short Exact Sequences
Last time we defined a short exact sequence in an abelian category 
to be an exact sequence of the form 
. These are the objects of a category 
, whose morphisms are triples of arrows 
, 
, 
making the two squares commute in the diagram:
The category is clearly enriched over 
.
As a first step into this category, we show that if 
and 
are both monic, then 
is monic as well. First let’s see how to do this in . We want to show that if 
for 
that 
. Now 
, but we assumed to be monic, so 
— 
. Since the top row is exact this means 
, and so there’s an 
so that 
. Now 
, but both 
and are monic. Thus 
, and 
as well.
That’s the normal way to proceed, and we call it a “diagram chase”. Just start with an element on the diagram at 
and “chase” it around the diagram. But this only works if is made up of structured sets, and we haven’t assumed that at all! We need to work a tiny bit more abstractly and look only at the arrows.
So let’s take 
. Now 
. Since is monic, we have 
. And thus 
factors uniquely through 
as 
. Now 
. Since and are assumed monic, 
, and so 
and is monic.
Either way, we can dualize the whole theorem to say that if and are epic, then is epic too. Together, these results are called the “short five lemma”.
Here’s another lemma, based on this diagram:
The square on the right is a pullback, and I say that if is epic then 
is too. On the left is the kernel of , and I say that it factors through the kernel 
of to make the diagram commute.
Let’s just skip the chase and go directly to the general picture. As usual, pullbacks can be defined by products and equalizers. We get the product as 
, which then has two arrows to 
: 
and 
. And we find their equalizer by taking the kernel of their difference: 
. Then 
and 
.
Now under the assumption on , we know that 
is epic. Indeed, if 
then 
. But since is assumed epic, 
. Supposing now that 
, we see that 
, so 
factors through the cokernel of 
, which is . That is, 
, and so 
. Since is epic, 
, and so 
. Thus is epic.
As for the other assertion, the pair of arrows 
and 
satisfy 
, and so there is a unique arrow 
by the universal property of the pullback. In particular, 
. On the other hand, given any other 
with 
we have 
, so 
factors uniquely through as 
. Then 
and 
. But since the arrow factoring through a pullback is unique we must have 
. So really is the kernel of , as asserted.
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September 27, 2007 -
Posted by John Armstrong |
Category theory
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