# 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 | 8 Comments