# The Unapologetic Mathematician

## Homology

Today we can define homology before I head up to the Baltimore/DC area for the weekend. Anyone near DC who wants to hear about anafunctors can show up at George Washington University’s topology seminar on Friday.

As a preliminary, we need to know what quotients in an abelian category are. In $\mathbf{Ab}$ we think of an abelian group $G$ and a subgroup $H$ and consider two elements of $G$ to be equivalent if they differ by an element of $H$. This causes problems for us because we don’t have any elements to work with.

Instead, remember that $H$ comes with an “inclusion” arrow $H\rightarrow G$, and that the quotient has a projection arrow $G\rightarrow G/H$. The inclusion arrow is monic, the projection is epic, and an element of the quotient is zero if and only if it comes from an element of $G$ that is actually in $H$. That is, we have a short exact sequence $\mathbf{0}\rightarrow H\rightarrow G\rightarrow G/H\rightarrow\mathbf{0}$. But we know in any abelian category that this short exact sequence means that the projection is the cokernel of the inclusion. So in general if we have a monic $m:A\rightarrow B$ we define $B/A=\mathrm{Cok}(m)$.

Now we define a chain complex in an abelian category $\mathcal{C}$ to be a sequence $\cdots\rightarrow C_{i+1}\rightarrow C_i\rightarrow C_{i-1}\rightarrow\cdots$ with arrows $d_i:C_i\rightarrow C_{i-1}$ so that $d_{i-1}\circ d_i=0$. In particular, an exact sequence is a chain, since the composition of two arrows in the sequence is the zero homomorphism. But a chain complex is not in general exact. Homology will be the tool to measure exactly how the chain complex fails to be exact.

So let’s consider the following diagram

where $g\circ f=0$. We can factor $f$ as $m\circ e$ for an epic $e$ and a monic $m=\mathrm{Im}(f)$. We can also construct the kernel $\mathrm{Ker}(g)$ of $g$. Now $g\circ m\circ e=g\circ f=0=0\circ e$, so $g\circ m=0$ because $e$ is epic. This means that $m$ factors through $\mathrm{Ker}(g)$, and the arrow $\mathrm{Im}(f)\rightarrow\mathrm{Ker}(g)$ must be monic.

Now, if the sequence were exact then $\mathrm{Im}(f)$ would be the same as $\mathrm{Ker}(g)$, and the arrow we just constructed would be an isomorphism. But in general it’s just a monic, and so we can construct the quotient $\mathrm{Ker}(g)/\mathrm{Im}(f)$. When the sequence is exact this quotient is just the trivial object $\mathrm{0}$, so the failure of exactness is measured by this quotient.

In the case of a chain complex we consider the above situation with $f=d_{i+1}$ and $g=d_i$, so they connect through $C_i$. We define $Z_i=\mathrm{Ker}(g)$ and $B_i=\mathrm{Im}(f)$, which are both subobjects of $C_i$. Then the “homology object” $H_i$ is the quotient $H_i=Z_i/B_i$. We can string these together to form a new chain complex $\cdots\rightarrow H_{i+1}\rightarrow H_i\rightarrow H_{i-1}\rightarrow\cdots$ where all the arrows are zero. This makes sense because if we think of the case of abelian groups, $H_i$ consists of equivalence classes of elements of $Z_i$, and when we hit any element of $Z_i$ by $d_i$ we get ${0}$. Thus the residual arrows when we pass from the original chain complex to its homology are all zero morphisms.

October 3, 2007 - Posted by | Category theory

1. Yay! Now we’re talking! :)

Seriously, reading these posts of yours makes me think about whether it’d be reasonably easy to feed this kind of homological algebra into Coq, or even, possibly, automate diagram chasing proofs with proof assistants.

Comment by Mikael Johansson | October 4, 2007 | Reply

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