Diagram Chases Done Right
Blatantly stealing my title from Axler’s (in)famous book, I’m going to take the lemmas from last time and show that we actually can prove things in arbitrary abelian categories using “diagram chases”.
Remember that this technique started in categories of modules where the objects are actually structured sets with elements, but we don’t want to assume that our objects are like this. When we were dealing with enriched categories, we talked about elements of the underlying set by looking at arrows from the monoid identity object, but we don’t have that floating around either, since our abelian category might not be monoidal.
Instead, we’ll generalize even further to “members” of an object. We call any arrow with codomain
a member of
and write
. Now we say that two such arrows are equivalent — and write
if there are epimorphisms
and
with
. Clearly the relation is reflexive (use the identity on the source of
) and symmetric. We can see transitivity by considering the following diagram:
We assume that there are epimorphisms and
in the upper-right so that
, and epimorphisms
and
in the lower-left so that
. Now we can pull back the upper left square to get
and
, which are epic by the second lemma last time. The compositions
and
are then the epimorphisms we need to show that
.
The members of behave sort of like an abelian group in that we have a “zero” member
, and each arrow has a negative
in its hom-set. It’s straightforward to show that if
then
and any zero morphism
is equivalent to the zero member. Also, any arrow
sends a member
to a member
, so it behaves like a function on sets.
Now we have the following elementary rules for diagram chases, intended to parallel the rules we’d have if we were actually talking about a category of modules:
is monic if and only if for all
,
implies
is monic if and only if for all
,
implies
is epic if and only if for all
there is an
with
is zero if and only if for all
,
- Given
and
, the sequence
is exact at
if and only if
, and for every
with
there is an
with
- Given
and
with
there is a
with
. Further, any
with
has
and any
with
has
.
The first two rules are just the definition of a monic arrow over again, and the fourth is just the definition of zero morphisms. For the third rule, if is epic then we can pull back
along
to find
and
with
, which implies
. Conversely, if
is not epic then the member
is not of the form
for any
.
Rule six seems complicated, but it just replaces subtraction of elements. Since there are epimorphisms
so that
. Then all the conclusions hold by taking
.
Finally, in rule five we can take the canonical factorization . If the sequence is exact at
then
is the kernel of
, so for any
with
we have
. Now we can pull back
along
to get
and
with
and
epic. We see that
, so
, as desired. Conversely, if this holds for all
we can take
the kernel of
so that
. Then there is some
with
, and thus there are epimorphisms
with
. This implies that
factors through
, and thus that
as subobjects of
. But
implies that
, and the sequence is exact.
We can use these rules to replace everything we’d normally want to do with elements of modules. Just write everything out as if you were looking at a category of modules so you can take elements and chase them around the diagram. Then replace “element of” with “member of”, “equals” with “equivalent”, and so on. The rules above show that the semantics of membership and equivalence in arbitrary abelian categories are exactly the same as those of elementhood and equality in the category of abelian groups, and so everything carries over.
This approach has a significant advantage over the other major approach, which is to show that every abelian category can be faithfully embedded as a subcategory of a module category. In that case we can work with the image of the embedding functor, where every object is a module and every morphism is a homomorphism, and the naïve semantics of diagram chases applies. However it takes all the machinery of an embedding theorem, which might be a ways off. Here, we don’t have to use anything nearly so fancy. We just say the exact same words as we would in , but give them a different meaning to apply to our different context.
Next time I’ll actually do a diagram chase with the new semantics to illustrate how smoothly it works.
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Hi! what do you mean, in the proof of the 5th point of the rule for diagram chasing, by “pulling back y’ along e to get e'” ? thanks in advance!
I mean to form the pullback square just like in the proof of the third rule.