Isomorphism between $H_k(X,x_0)$ and $\tilde H_k (X)$

Question

I can prove this using the long exact sequence of relative homology groups. Is there a way to understand this isomorphism intuitively?

Answer 1

Given $A\hookrightarrow_{j} X$ we have by definition $ C(X,A)=C(X)/C(A) $, from the long exact sequence and the group isomorphism theorems we get $H_n(X,A)=\frac{\{\alpha\in C_n(X)|\partial\alpha\in jC_{n-1}(A)\}}{\partial C_{n+1}(X)+C_n(A)}$. In this way you can have a more concrete way to understand the homology of the pair.

Answer 2

Relative homology $H_k(X,A)$ is when you allow cycles to have boundaries in $A$ (rather than requiring the boundary to be $0$), and when a cycle is nullhomologous if it is homologous to something in $A$. When $A$ is a single point, things become simpler.

When $k>0$, a cycle whose boundary is contained in a point has a degenerate boundary, which is homologous to $0$, so $H_k(X,x_0)$ is the same as $H_k(X)$ and $\widetilde{H}_k(X)$. (The latter is because reduced homology has the same chain groups as $H_k(X)$ except at the $-1$ grading.)

The interesting case is $k=0$. $0$-cycles are points, and two points are homologous if there is a path between them. And, in relative homology, a point is nullhomologous if there is a path to $x_0$. This kills one of the connected components. $\widetilde{H}_0(X)$ kills off a path component in a slightly different way, which is by saying that the sum of the coefficients of the points must be zero. The map $\widetilde{H}_0(X)\to H_0(X,x_0)$ can be thought of as setting the coefficients of points in the same path component of $x_0$ to $0$. The inverse to this is to sum the coefficients of the points in a cycle representative for $H_0(X,x_0)$, then subtract that amount of $x_0$.