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In Hatcher, a Klein Bottle $K$ is considered to be two Möbius band $A,B$ glued together. I see the map $\phi$ is $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$, but why $1 \mapsto (2,-2)$? I other words, how can I get the conclusion that the 1 on the boundary maps to different signs on the boundary?

Can orient the second Möbius band such that $1$ on the intersection map to $(2,2)$? It is just matter of orientation right?

The map $\phi$ is $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}, 1 \mapsto (2,-2)$: $$0 \to H_2(K) \to H_1(A \cup B) \stackrel{\phi}{\to} H_1(A) \oplus H_1(B) \to H_1(K) \to 0$$

azimut
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1LiterTears
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  • What is the map $\phi$? is it the map in the Mayer-Vietoris sequence? – Moss Nov 13 '13 at 03:31
  • Yes, @Sebastian – 1LiterTears Nov 13 '13 at 03:31
  • If you read the construction of the Mayer-Vietoris sequence in Hatcher (page 149-150), the map $\phi$ comes from a map $\varphi$ on chains. Now the map $\varphi$ on chains sends $x\mapsto (x,-x)$. The reason for mapping like is so that one gets a short exact sequence of chain complexes, hence a long exact sequence in homology (the Mayer-Vietoris sequence) – Moss Nov 13 '13 at 03:38
  • Thank you@Sebastian. Here what I don't understand is that on the top of P151, it says that $H_1(K) \approx \mathbb{Z} \oplus \mathbb{Z}_2$ since we can choose (1,0), (1,-1) as basis for \mathbb{Z} \oplus \mathbb{Z}$. I don't understand this, that how do we determine how many basis? – 1LiterTears Nov 13 '13 at 12:58
  • Hi I asked here @Sebastian - hope you may have chance to take a look at it? http://math.stackexchange.com/questions/566271/hatcher-p151-example-2-47 – 1LiterTears Nov 14 '13 at 03:38
  • I posted an answer to your question, I hope it helps. – Moss Nov 14 '13 at 05:47
  • @Moss Why $(2,-2)$ though and not $(1,-1)$? It says: $x \mapsto (x,-x)$. – Anon Jan 08 '23 at 17:19

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