Isomorphism $H_n(D^n, S^{n - 1};A)$ and $A$ The homomorphism $a \mapsto [[a \cdot f]]$ is supposed to be an isomorphism between $A$ and $H_n(D^n, S^{n - 1};A)$ for any homeomorphism $f: \Delta^n \to D^n$. I have proven that it is well defined and injective, but why is it surjective?
The group $H_n(D^n, S^{n - 1};A)$ is generated by elements of the form $a \cdot \sigma + b \cdot \tau + c \cdot \partial(\kappa)$ where $\sigma: \Delta^n \to D^n$, $\tau: \Delta^n \to S^{n - 1}$ and $\kappa: \Delta^{n + 1} \to D^{n + 1}$ and I do not see how an element of this form represents the same equivalence class of $[[m \cdot f]]$ for some $m \in A$. We can write $\sigma$ as $f f^{-1} \sigma$ to get $f$ in there, but I don't see that working out. Any hints?
 A: Once you have enough machinery set up, an overriding theme of singular homology in topology is this: Avoid computing anything with it on the level of singular chains! 
In general, it seems to me that only in very simple circumstances can you hope to say something concrete using singular chains. That's not to say one should never use singular chains, but that singular homology is mostly useful as a tool for showing that different methods for computing homology groups agree and satisfy the various properties you would hope they have as well as providing a concrete geometric interpretation for what all these theories attempt to capture (a kind of extension problem through bordism of triangulated manifolds with cone-like degeneracies whose simplices are coherently oriented).
I'm going to assume that $A$ is a ring first and then comment on what needs to change if $A$ is an $R$-module (e.g., an abelian group).

Here are some hints for how I would approach this with $A$ a ring.


*

*$f$ restricts to a homeomorphism $\partial\Delta^n\cong S^{n-1}$. This is the ''invariance of boundary'' for topological manifolds with boundary. You can do it in this special case with the standard trick of deleting a point. If $x\in \partial\Delta^n$, then $f\colon \Delta^n\setminus\{x\}\cong D^n\setminus\{f(x)\}$. So if $f(x)\notin S^{n-1}$...

*This implies that $f_\ast\colon H_n(\Delta^n,\partial\Delta^n;A)\approx H_n(D^n,S^{n-1};A)$. 

*Show that $a\mapsto [a\cdot id_{\Delta^n}]$ is an isomorphism $A\to H_n(\Delta^n,\partial \Delta^n;A)$. There are two ways to do this.


*

*If you are comfortable with simplicial homology use the relative version of it to deduce that $H_n(\Delta^n,\partial \Delta^n;A)$ is generated by $[id_{\Delta^n}]$ as a free $A$-module. 

*Argue by induction where $n=0$ is basically automatic. I remember that Hatcher actually gives this sort of argument in his book and uses it to prove that simplicial and singular homology agree. In his book, it is Example 2.23. You'll have to unpack the boundary map in the LES to do this.




Now I'll assume $A$ is an $R$-module (e.g., an abelian group). Note that the equivalence of singular and simplicial homology goes through with any coefficient module. The only thing you need to know is that if $A$ is an $R$-modules (e.g., an abelian group) then the last step above is true in the sense that every element of $H_n(\Delta^n,\partial\Delta^n;A)$ has the form $[a\cdot id_{\Delta^n}]$. You can prove this as Hatcher does in Example 2.23 with a simple modification of the proof. The proof of the equivalence then goes through mutatis mutandis.
With that said, it is clear that if you replace $A$ by an $R$-module above (e.g., an abelian group) then bullet point three remains the same. There are again two ways to prove this last item in the case $A$ is a module.


*

*Use relative simplicial homology to show that every element of $H_n(\varDelta^n,\partial\varDelta^n;A)$ has the form $[a\cdot id_{\Delta^n}]$.

*Explicitly write out the proof of the assertion in the first paragraph under the second horizontal line (i.e., the first paragraph under the horizontal line above).

