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My question about the first homology group of the Klein bottle via Mayer-Vietoris sequence. I have the exact sequence below

$ 0 \xrightarrow{} H_1( S^1) \xrightarrow{(i,j)} H_1(M) \oplus H_1(M^\prime )\xrightarrow{k-l} H_1(K) \xrightarrow{\partial_1} \tilde{H_0} = 0 $ where $M$ and $M^\prime$ are both Möbius strip. I have also the central map

$(i,j)$ sends $1$ to $(2,−2)$. What is the mathematical explanation of

$(i,j)$ sends $1$ to $(2,−2)$

I see explanation "Since the boundary circle of a Möbius band wraps twice around the core circle" everywhere.

But I can't see why we care about core circle ? And I want to see the first homology group as a continuous function from $[0,1]$ to my space. How can I see this central map from this perspective?

Please, give me hints firstly, rather than precise answer.

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The Mayer-Vietoris sequence is defined as the long exact sequence

$$H_*(U \cap V) \to H_*(U) \oplus H_*(V) \to H_*(X)$$

where the first map is defined by $(i_*, j_*)$ and the second map is defined by $k_* - l_*$ where $i, j$ are the inclusion maps of $U \cap V$ inside $U$ and $V$ and $k, l$ are the inclusions of $U$ and $V$ inside $X$.

If $X$ is the Klein bottle, $\{U, V\}$ cover of it by the two mobius strips, then $U \cap V$ is homotopy equivalent to the boundary of the two strips. I have to thus compute the maps $i_*, j_*$ induced on homology where $i, j$ are inclusions of the boundary map of the mobius strips into the Klein bottle.

That's precisely where we need the wrapping description. If $M$ is a moebius strip then the inclusion map $\partial M \hookrightarrow M$ could be understood as follows: deformation retract $M$ to it's core circle, which generates $H_1(M)$. Under this deformation retract $\partial M$ maps to twice the core circle. So $i_* : H_1(\partial M) \to H_1(M)$ is precisely the multiplication by $2$ map.

Thus, $i_*$ is multiplication by $2$ whereas $j_*$ is multiplication by $-2$ due to orientation issues. Tat tells you $\alpha$ sends $1$ to $(-2, 2)$ once you identify $H_1(U \cap V)$ with $\Bbb Z$ and $H_1(U) \oplus H_1(V)$ with $\Bbb Z \oplus \Bbb Z$.

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  • $\begingroup$ Thanks but how can you said deformation retract of M to its core circle generates H_1(M) ? $\endgroup$ – rmznyzgyr Apr 18 '16 at 21:02
  • $\begingroup$ @rmznyzgyr Mobius strip $M$ deformation retracts to it's core circle, $H_1$ of which is $\Bbb Z$ and it's generated by the (homology class of the) circle itself. I am not sure what's unclear about this. $\endgroup$ – Balarka Sen Apr 18 '16 at 21:06

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