A question on cell attachments. Let $n \geq 2$ and let $\alpha: S^{n-1} \to S^{n-1}$ be a continuous map of degree $a$, meaning that the homomorphism induced on the $n$-th homology is multiplication by $a$. Let $S^{n-1} \cup_{\alpha} D^n$ be the space obtained by attaching an $n$-dimensional cell to $S^{n-1}$ along $\alpha$. Compute $H_k( S^{n-1} \cup_{\alpha} D^n)$ for all $k$.
The homology groups I am talking about here is the ones of the complex $S_{\ast}(X)$ for a topological space $X$, more precisely  $S_{\ast}(X)$ is the complex $... \to S_n(X) \to ... \to S_2(X) \to S_1(X) \to S_0(X)$ where $S_k(X)$ is the free abelian group generated by all singular simplexes of dimension $k$, i.e all continuous maps $\sigma: \Delta^k \to X$. Kind of feel lost on this exercise, and it is supposed to be an easy one...
 A: The Mayer-Vietoris theorem says we have a long exact sequence 
$$
0\to H_n(S^{n-1} \cup_{\alpha} D^n)\overset{\partial}\to H_{n-1}(S^{n-1} \cap D^n)\overset{i_1\oplus i_2}\to H_{n-1}(S^{n-1})\oplus H_{n-1}(D^n)\overset{p_1-p_2}\to H_{n-1}(S^{n-1} \cup_{\alpha} D^n)\to0
$$
where $i_1,i_2,p_1,p_2$ are the subspace inclusion maps, respectively, $S^{n-1}\cap D^n\hookrightarrow S^{n-1},S^{n-1}\cap D^n\hookrightarrow D^n,S^{n-1}\hookrightarrow S^{n-1} \cup_{\alpha} D^n,$ and $D^n\hookrightarrow S^{n-1} \cup_{\alpha} D^n.$ Then from the induced maps in homology are formed $i_1\oplus i_2$ and $p_1-p_2.$
And we have $S^{n-1}\cap D^n\simeq S^{n-1},$ $H_{n-1}(D^n)\cong0,$ $i_2=0=p_2,$ and $i_1=\alpha$ is the degree $a$ map. And of course $H_{n-1}(S^{n-1})\cong\mathbb{Z}.$
Since $\alpha$ is injective, we have $H_n(S^{n-1} \cup_{\alpha} D^n)\cong 0.$ 
Now our sequence is $$0\to \mathbb{Z}\overset{a}\to\mathbb{Z}\to H_{n-1}(S^{n-1} \cup_{\alpha} D^n)\to 0,$$ from which it follows $H_{n-1}(S^{n-1} \cup_{\alpha} D^n)=\mathbb{Z}/a\mathbb{Z}.$
The part which I think is a little confusing is, how can $i_1$ be the non-injective degree $a$ map, when it is supposed to be the subspace inclusion 
$S^{n-1}\cap D^n\cong S^{n-1}\hookrightarrow S^{n-1}$? Surely this inclusion is just the identity map?
Well, no. Remember that Mayer-Vietoris requires that we use open sets $A$, $B$ that cover the total space $X=S^{n-1} \cup_{\alpha} D^n$. $S^{n-1}$ and $D^n$ do not do this, they are closed sets, $S^{n-1}$ contains no neighborhoods of $X$. So we must fatten up our sphere, and retract our disk a bit. Something like $A=S^{n-1}\times[0,\frac{2}{3})$ and $B=S^{n-1}\times(\frac{1}{3},1]/S^{n-1}\times \{1\}$ (so a cone on $S^{n-1}$ that doesn't go "all the way out", so to speak). And although they are homotopy equivalent to $S^{n-1}$ and $D^n$, respectively, it is these subspaces $A$ and $B$ which should have appeared in the MV sequence above, and which determine the maps.
Then we see the intersection $A\cap B\hookrightarrow A$ does map into the fat boundary via the degree $a$ attaching map. 
See also this answer.
