Relative homology of the Klein bottle Let $K$ denote the Klein bottle. You may assume without proof that $K$ is the union of two copies $M$ and $N$ of the Moebius band with boundaries glued by a homeomorphism. Compute $H_1(K; M)$ and $H_2(K; M)$.
I tried using relative homology and it doesn't work out well.  Any suggestions?  Thanks.
 A: Hint: excise an annulus $m$ (a fattening of the central circle) from the Moebius band $M$, then $H_*(K\setminus m, M\setminus m)\simeq H_*(K,M)$ by excision. There is a homotopy equivalence of pairs $(K\setminus m, M\setminus m)\simeq (N,\partial N)$, and the homology of the pair will be calculable using the long exact homology sequence of that pair, with both $N$ and $\partial N$ homotopy equivalent to a circle.
To be precise, there are pretty self evident homotopy equivalences $\partial N\xrightarrow{\sim} S^1$ and $N\xrightarrow{\sim} S^1$ such that the diagram below commutes
\begin{array}{ccc}
\partial N &\hookrightarrow &N\\
\;\;\downarrow\sim&&\;\;\downarrow\sim\\
S^1&\xrightarrow{z\mapsto z^2}&S^1
\end{array}
So that in homology the inclusion $\partial N\hookrightarrow N$ induces an isomorphism $H_0(\partial N)\xrightarrow{\sim}H_0(N)$ and a map $H_1(\partial N)\rightarrow H_1(N)$ that is equivalent to $\Bbb Z\xrightarrow{2\times-}\Bbb Z$, so that from the long exact sequence of the triple $(N,\partial N,\emptyset)$, which here reduces to
$$
\begin{array}{cccc}
0\to H_0(\partial N) \xrightarrow{\sim} H_0(N) \to H_0(N,\partial N) \xrightarrow{\partial} &H_1(\partial N)&\rightarrow&H_1(N)&\to H_1(N,\partial N)\xrightarrow{\partial}0\\
&\;\;\downarrow\sim&&\;\;\downarrow\sim&\\
&H_1(S^1)&\xrightarrow{i_2}&H_1(S^1)&\\
&\;\;\downarrow\sim&&\;\;\downarrow\sim&\\
&\Bbb Z&\xrightarrow{2\times -}&\Bbb Z&
\end{array}
$$
From this it follows that
$$H_*(K,M)\simeq H_*(N,\partial N)\simeq\begin{cases}\Bbb Z/2\Bbb Z&\text{if }*=1\\0&\text{otherwise} \end{cases}\;.$$
