# Computing homology w/ Mayer Vietoris

Let $(H_*, \partial_*)$ be a homology theory satisfying the dimension axiom. Let $n\ge1$ and $X= S^n \cup_f D^{n+1}$ where $f:S^n\to S^n$ is a degree $k$ map. Compute each homology group of $X$.

I have computed all homology groups, except $H_0(X)$. To compute $H_0(X)$, consider this part of the Mayer Vietoris sequence:

$...\to \underbrace{H_0(S^n)}_{\Bbb{Z}} \to \underbrace{H_0(S^n)\oplus H_0(D^{n+1})}_{\Bbb{Z}\oplus \Bbb{Z}} \to H_0(X) \to 0 \to ...$

where the first map is $(H_0(f), H_0(i))$. I expect this map to be $x\mapsto (x,x)$ up to isomorphism, but I haven't been able to prove it. Please give me a hint.

• Watch out for the case $n=0$. Jun 2, 2015 at 11:45
• Sorry, forgot to mention: $n \geq 1$ Jun 2, 2015 at 12:58

This works for every base ring $R$ as follows:
Choose point inclusions $\iota_1\colon*\rightarrow S^n$ and $\iota_2\colon*\rightarrow S^n$, such that $f\circ\iota_1=\iota_2$. Now argue that both point inclusions induce an iso on $H_0$. (Do you need help to manage that?) Now fix an iso $H_0(*)\cong R$. With respect to these chosen isos the first component of the map $H_0(S^n)\rightarrow H_0(S^n)\oplus H_0(S^n)$ is the identity on $R$. Almost the same reasoning works for the second factor and you will obtain your expected result.
• It's not clear to me why the inclusions induce isos on $H_0$. Can you please elaborate? Jun 2, 2015 at 13:44
• Of course. Since you already know that $H_0(S^n)\cong R$ holds, the map is up to iso a self map of $R$, hence a multiplication with an element of the ring. By the same reasoning, the left inverse (induced by the constant map on the level of spaces) is the multiplication by another element of the ring. Since the ring is commutative, you can conclude that those two elements are inverse to each other, hence the maps are mutually inverse isos. Jun 2, 2015 at 13:52
• This works also for $H_0(D^{n+1})$ as long as you know that $H_0(D^{n+1})\cong R$ holds. Jun 2, 2015 at 13:53