Question about Möbius strip relative homology group I need some clarification about the first homology group of the couple (Möbius strip, Border).
In my exercises, I use the theorem that says that the succession of
$$0\to H_1(\partial M)\to H_1(M)\to H_1(M,\partial M)\to 0$$
is an exact one.
If $f$ is the function between $M$ and $\partial M$ I can say that the generator of $\partial M$ got multiplied by two, so $H_1(M)\simeq H_1(\partial M)\simeq \mathbb{Z}$ but those two homology group are not isomorphic.
What I don't understand is why $H_1(M,\partial M)\simeq$coker($f$), can anyone help me with that passage?
 A: So, here $f\colon M\hookrightarrow \partial M$ is the inclusion map. And using long exact sequence for the pair $(M,\partial M)$ we have $$\cdots\longrightarrow H_2(M,\partial M)$$$$\longrightarrow H_1(\partial M)\xrightarrow{f_*}H_1(M)\xrightarrow{i_*} H_1(M,\partial M)$$$$\longrightarrow H_0(\partial M)\xrightarrow{f_*}H_0(M)\longrightarrow\cdots$$ Here, $i:(M,\varnothing)\hookrightarrow (M,\partial M)$ is also the inclusion map.
Now, $f_*\colon \Bbb Z\cong H_1(\partial M)\to H_1(M)\cong \Bbb Z$ is multiplication by $2$ due to the $2$-fold covering map $\partial M\to \text{Central Circle}$. In particular, $f_*\colon H_1(\partial M)\to H_1(M)$ is injective.
Also, $\partial M$ and $M$ both are path-connected so, $f_*\colon H_0(\partial M)\to H_0(M)$ is an isomorphism.

Answer To Your Question: So,
we have a short exact sequence $$0\longrightarrow H_1(\partial M)\xrightarrow{f_*}H_1(M)\xrightarrow{i_*} H_1(M,\partial M)\longrightarrow0.$$ Now, $$\text{coker}\left(H_1(\partial M)\xrightarrow{f_*}H_1(M)\right)=\frac{H_1(M)}{\text{im}\left(H_1(\partial M)\xrightarrow{f_*}H_1(M)\right)}$$$$=\frac{H_1(M)}{\ker\left(H_1(M)\xrightarrow{i_*} H_1(M,\partial M)\right)}\cong H_1(M,\partial M).$$ Here, the first equality is the definition of co-kernel. Second equality is due to short exact sequence, actually exactness at $H_1(M)$. The last isomorphism is due to the First Isomorphism Theorem for groups.
