# The Unapologetic Mathematician

## 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 $x$ with codomain $A$ a member of $A$ and write $x\in_mA$. Now we say that two such arrows are equivalent — and write $x\equiv y$ if there are epimorphisms $u$ and $v$ with $x\circ u=y\circ v$. Clearly the relation is reflexive (use the identity on the source of $x$) and symmetric. We can see transitivity by considering the following diagram:

We assume that there are epimorphisms $t$ and $u$ in the upper-right so that $x\equiv y$, and epimorphisms $v$ and $w$ in the lower-left so that $y\equiv z$. Now we can pull back the upper left square to get $u'$ and $v'$, which are epic by the second lemma last time. The compositions $t\circ u'$ and $w\circ v'$ are then the epimorphisms we need to show that $x\equiv z$.

The members of $A$ behave sort of like an abelian group in that we have a “zero” member $\mathbf{0}\rightarrow A$, and each arrow has a negative $-x$ in its hom-set. It’s straightforward to show that if $x\equiv y$ then $-x\equiv -y$ and any zero morphism $0:B\rightarrow A$ is equivalent to the zero member. Also, any arrow $f:A\rightarrow B$ sends a member $x\in_mA$ to a member $(f\circ x)\in_mB$, 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:

• $f:A\rightarrow B$ is monic if and only if for all $x\in_mA$, $f\circ x\equiv0$ implies $x\equiv0$
• $f:A\rightarrow B$ is monic if and only if for all $x,x'\in_mA$, $f\circ x\equiv f\circ x'$ implies $x\equiv x'$
• $f:A\rightarrow B$ is epic if and only if for all $y\in_mB$ there is an $x\in_mA$ with $f\circ x\equiv y$
• $f:A\rightarrow B$ is zero if and only if for all $x\in_mA$, $f\circ x\equiv0$
• Given $f:A\rightarrow B$ and $g:B\rightarrow C$, the sequence $A\rightarrow B\rightarrow C$ is exact at $B$ if and only if $g\circ f=0$, and for every $y\in_mB$ with $g\circ y\equiv0$ there is an $x\in_mA$ with $f\circ x\equiv y$
• Given $f:A\rightarrow B$ and $x,y\in_mA$ with $f\circ x\equiv f\circ y$ there is a $z\in_mA$ with $f\circ z\equiv0$. Further, any $g:A\rightarrow C$ with $g\circ x\equiv0$ has $g\circ y\equiv g\circ z$ and any $h:A\rightarrow D$ with $h\circ y\equiv0$ has $h\circ x\equiv -h\circ z$.

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 $f$ is epic then we can pull back $y$ along $f$ to find $f'$ and $x$ with $f\circ x=y\circ f'$, which implies $y\equiv f\circ x$. Conversely, if $f$ is not epic then the member $1_B\in_mB$ is not of the form $f\circ x\equiv1_B$ for any $x\in_mA$.

Rule six seems complicated, but it just replaces subtraction of elements. Since $f\circ x\equiv f\circ y$ there are epimorphisms $u,v$ so that $f\circ x\circ u=f\circ y\circ v$. Then all the conclusions hold by taking $z=y\circ v-x\circ u\in_mA$.

Finally, in rule five we can take the canonical factorization $f=m\circ e$. If the sequence is exact at $B$ then $m$ is the kernel of $g$, so for any $y$ with $g\circ y=0$ we have $y=m\circ y'$. Now we can pull back $y'$ along $e$ to get $e'$ and $x\in_mA$ with $y'\circ e'=e\circ x$ and $e'$ epic. We see that $y\circ e'=f\circ x$, so $y\equiv f\circ x$, as desired. Conversely, if this holds for all $y\in_mB$ we can take $k\in_mB$ the kernel of $g$ so that $g\circ k\equiv0\in_mC$. Then there is some $x\in_mA$ with $f\circ x\equiv k$, and thus there are epimorphisms $u,v$ with $k\circ u=m\circ e\circ x\circ v$. This implies that $k$ factors through $m$, and thus that $\mathrm{Ker}(g)\leq\mathrm{Im}(f)$ as subobjects of $B$. But $g\circ f=0$ implies that $\mathrm{Im}(f)\leq\mathrm{Ker}(g)$, 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 $\mathbf{Ab}$, 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.

September 28, 2007 - Posted by | Category theory

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