Exact sequence of a pair: Computing the connecting map

Given a pair of spaces $(X,A)$ , with $A \subset X$ , the short exact sequence of the pair is a relationship between the chains $C_*(A),C_*(X/A)$ and $C_*(X)$ . The sequence that looks like this:

(1) $\begin{equation*} 0 \longrightarrow C_*(A) \xlongrightarrow{i} C_*(X) \xlongrightarrow{j} C_*(X/A) \longrightarrow 0 \end{equation*}$

is short exact, or that: $\Ker{j} = \Im{i}$ . If you want to know more about exact sequences I recommend reading this.

It turns out that on the level of homology this short exact sequence turns into a long exact sequence:

(2) $\begin{equation*} \ldots \longrightarrow H_*(A) \xlongrightarrow{i_*} H_*(X) \xlongrightarrow{j_*} H_*(X/A) \xlongrightarrow{\delta} H_{*-1}(A) \ldots \end{equation*}$

and the map $\delta$ is called the connecting homomorphism.

The natural question to ask is: What is $\delta([x])$ ? The answer is that
that $\delta([x]) = [\partial[x]]$ . However, if you read the aforementioned authoritative text on the subject you will see that $\delta$ is not explicitly defined, but its existence is merely proved.

However, given a particular $X$ and $A$ , one is able to come up with an idea of what the map does. I’ll go over the basic diagram chase and then show how this works by example.

Since we are given a representation of $H_*(X/A)$ we can begin by taking a homology class $\hat{c}$ which is relative cycle. Any chain $c$ in $X$ which is in the class $\hat{c}$ has the property that $j(c) = \hat{c}$ in $H_*(X/A)$ . Such a chain $c$ is called a lift.

Assuming we have found a lift we can compute $z = \partial(c)$ , and it turns out that $z \in Z_{*-1}(A)$ (because $\partial{j} = j\partial$ ), which means that $[z] \in H_{*-1}(A)$ . So our mapping sends $\hat{c} \in H_*(X/A)$ to $[z] \in H_{*-1}(A)$ . So you now just need to prove that this process is well defined and this ends your diagram chase as well as your interest in the details of this sequence. Unless of course you are Mr. Cooperman, or, a computer scientist.

If you are a computer scientist you might be dissatisfied because it is not entirely clear how $[z]$ is presented. Indeed $z$ itself is not unique as it depends on our choice of lift, $z$ is likely itself not even a member of our given representative for homology on $H_{*-1}(A)$ .

The problem we have is that so far our cycle $z$ is written down as a linear combination of cells. We want an equivalent representation in terms of cycles. Noether says we can write down $Z_{*}(A) \simeq H_{*}(A) \oplus B_{*}(A)$ .
This tells us that we can use our homology basis to form a basis for our cycle group, and its not to hard to see that we can extend the cycle group to a basis for all chains. However in our case we want to take a cycle expressed in the chain basis, and write it as a linear combination of given cycles. This amounts to being able to write down a rectangular “change of basis” matrix between the cell basis and the given cycle basis on $Z_{*-1}(A)$ .

The column space of this matrix would be given by our given cycle basis, and the row space would be given by the canonical cell basis for $C_{*-1}(A)$ . We then look to “solve $Ax = z$ ” over $\mathbb{Z}_p$ (in this blog $p = 2$ ), in the sense that we want to write down a linear combination of the cycles in $Z_{*-1}(A)$ apply our matrix and end up with the $z$ we computed.

Consider this space $A$ :

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which is homotopic to an annulus.

We can imagine that $A \subset X$ , and that $X$ is obtained by gluing a disc onto the boundary of $A$ , so that $X$ is itself homotopic to a disc, $X/A$ is then homotopic to a sphere, with exactly one non-trivial 2nd homology class. The boundary of this class would be the outermost cycle of $A$ .

Lets say that we are provided a basis for $H_1(A) = \langle \alpha = bc + cd + db \rangle$ and $B_1 = \langle \beta = ab + bc + ca \rangle$ . If I want to express the outer bounding cycle $ab + bd + dc +ca$ in this basis, I begin by writing down the change of basis matrix, in order to solve $Ax=z$ I augment the matrix with the identity matrix on the right and $z$ on the bottom.

$A' = \left( \begin{array}{c|ccccc|cc} & ab & ac & bc & bd & cd & \alpha & \beta \\ \hline \alpha & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\ \beta & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\ \hline & 1 & 1 & 0 & 1 & 1 & 0 & 0 \end{array} \right)$

Now you can see that modulo our extra row for $z$ our matrix is in row echelon form, and it suffices to perform one Gaussian elimination (over $Z_2$ ) in this last column to achieve our result:

$\left( \begin{array}{c|ccccc|cc} & ab & ac & bc & bd & cd & \alpha & \beta \\ \hline \alpha & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\ \beta & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\ \hline z & 1 & 1 & 0 & 1 & 1 & 0 & 0 \end{array} \right)$

$z \rightarrow z + \beta$

$\left( \begin{array}{c|ccccc|cc} & ab & ac & bc & bd & cd & \alpha & \beta \\ \hline \alpha & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\ \beta & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\ \hline z=\beta & 0 & 0 & 1 & 1 & 1 & 0 & 1 \end{array} \right)$

$z \rightarrow z + \alpha$

$\left( \begin{array}{c|ccccc|cc} & ab & ac & bc & bd & cd & \alpha & \beta \\ \hline \alpha & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\ \beta & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\ \hline z=\alpha+\beta & 0 & 0 & 0 & 0 & 0 & 1 & 1 \end{array} \right)$

Notice how the right hand side now contains our vector $x$ and indeed $Ax = z = \partial(j(c))$ .