Show that the identity map $\Delta^n \rightarrow \Delta^n$ is a basis for $H_n(\Delta^n, \delta\Delta^n; R)$.
Here $\Delta^n$ is the n-simplex, and I know $\delta \Delta^n$ denotes its "boundary", but I'm not completely sure what that means--I think it's the set of all n-1-simplices contained in $\Delta^n$, but I'm not sure. Most importantly, I don't understand what it means for a map to be a basis for a homology group. I wrote down the following anyway:
If n = 0, $\delta\Delta^0$ is empty and there is nothing to prove; if n = 1, $(\Delta^1, \delta\Delta^1)$ is a good pair and $\Delta^1/\delta\Delta^1$ is the circle $S^1$, and the identity map has only one element, which generates $\mathbb{Z}$. Otherwise, $H_n(\Delta^n; R) ≅ H_{n-1}(\Delta^n; R) ≅ 0$ since $\Delta^n$ is contractible, so the long exact sequence $... \rightarrow H_n(\delta\Delta^n; R)$ $\rightarrow H_n(\Delta^n; R)$ $\rightarrow H_n(\Delta^n, \delta\Delta^n; R)$ $\rightarrow H_{n-1}(\delta\Delta^n; R)$ $\rightarrow H_{n-1}(\Delta^n;R) \rightarrow ...$ has a section $0 \rightarrow H_n(\Delta^n, \delta\Delta^n; R)$ $\rightarrow H_{n-1}(\delta\Delta^n; R) \rightarrow 0$ and so $H_n(\Delta^n, \delta\Delta^n; R) ≅ H_{n-1}(\delta\Delta^n; R)$. The identity map takes each element of $\delta\Delta^n$ to itself, and $H_{n-1}(\delta\Delta^n)$ is generated by elements of $\delta\Delta^n$, so the result follows.
I am reasonably sure that this "proof" is wrong, probably for multiple reasons, but I don't really understand the question. I just tried to do what I know how to do to homology groups to come up with something that sounds vaguely related to the question. I want help understanding the question so I can make an actual answer to it, and I'm hoping what I wrote down will help clarify what I do and don't understand already about the question.