There is a sentence in Hatcher that I am really struggling to understand. We have the following theorem
If $X$ is a space and $A$ is a nonempty closed subspace that is a deformation retract of some neighborhood in $X$, then there is an exact sequence of reduced homology
$$\cdots\longrightarrow{H_n(A)}
\longrightarrow H_n(X)\longrightarrow H_n(X/A)\longrightarrow H_{n-1}(A)\longrightarrow H_{n-1}(X)\longrightarrow\cdots,$$
where the maps are induced by the inclusion $A\hookrightarrow X$ and the quotient map $X\rightarrow X/A$.
Let $\partial$ be the connecting homomorphism. There is then the phrase
The idea is that an element $x\in H_n(X/A)$ can be represented by a chain $\alpha$ in $X$ with $\partial\alpha$ a cycle in $A$ whose homology class is $\partial x\in H_{n-1}(A)$.
I am a little confused by what this means. It seems that the element $\partial\alpha$, where in this case I assume that the $\partial$ is the boundary map, must be an element of $H_{n-1}(A)$. But then why is $\alpha\in X$? I feel like it should be an element of $A$, so I'm a little confused by what's going on here.
Best Answer
$x$ lives in $X/A,$ but one could hope to express it as a $p \alpha,$ where $\alpha$ is lives in $X$ and $p$ is the projection $X \to X/A$. Then, since $\partial x = 0$ in $X/A,$ "on the level of $X$" $\partial x$ is also zero except some part of the boundary may be in $A,$ so that it becomes zero after projecting on $X/A.$ For example, for an interval $-$ $(X, A) = (I, \partial I)$, $I/\partial I$ is a circle, and the chain $\Delta^1 \xrightarrow{\simeq} I \to I/\partial I$ has zero boundary, so defines a homology class. And it can be considered as a chain in $I$ (the first arrow of the composition), whose boundary is then the difference between two points of $\partial I$. The substantial part of the statement is that any chain can be viewed like this.