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In Hatcher's Algebraic Topology Example 2.47, discussed in here and here, it states that for a Klein Bottle $K$, considered as two Möbius bands $A,B$ glued together we have:

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

Neither of the aforementioned posts clarifies, formally, why $1 \mapsto (2,-2)$ when the corresponding map between the chain groups is $x \mapsto (x,-x)$ (as stated in the book on p.150). Because of this, I would expect that $1 \mapsto (1,-1)$.

I understand that the boundary circle wraps twice around $A$ and $B$'s core circles, but I do not see how this formally leads to identifying that $1$ (which corresponds to the generator being the boundary circle(?)) gets mapped to $(2,2)$, especially in light of the $1 \mapsto (1,-1)$.

What am I missing?

John Palmieri
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Anon
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$A \cap B$ is the boundary of both $A$ and $B$, and it is homeomorphic to a circle. If we let $x$ be the cycle which goes once around this circle, then $x$ represents the generator of $H_1(A \cap B)$. Now if we include $A \cap B$ into $A$, then $x$ wraps twice around the Möbius band, and so $x$ maps to twice the generator: the map $A \cap B \to A$ induces the map $1 \mapsto 2$ on $H_1$.

Another way to say this is that $x$, viewed as a cycle in $A$, does not represent a homology generator but rather twice a homology generator.

John Palmieri
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  • Then the $x \mapsto (x,-x)$ is incorrect here and depends on the choice of $A,B$? I'm still having difficulty reconciling it with that statement. Doesn't the $1$ represent the generator (as you have stated)? If so - how can it be mapped on the one hand to $(1,-1)$ and on the other to $(2,-2)$? (I realize I'm misunderstanding something, but I cannot see what...). – Anon Jan 08 '23 at 18:16
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    @Anon Yes, the map is $x \mapsto (x, -x)$ but the image of the the cycle $x \in H_1(A\cap B)$ in $H_1(A)$ is twice the generator of $H_1(A)$. I think you are confusing the generator $1 \in H_1(A \cap B)$ with the generator $1 \in H_1(A)$. These $1$'s are misleading. These classes are represented by different cycles. – feynhat Jan 08 '23 at 19:20
  • As @feynhat says, and as my second paragraph says, $x$ represents "1" in $H_1(A \cap B)$ but represents "2" in $H_1(A)$, so the map can be described as both $x \mapsto (x, -x)$ and $1 \mapsto (2, -2)$. – John Palmieri Jan 08 '23 at 19:32
  • @feynhat I think I see it now - it's indeed $x, -x$, just we want $x$ to be stated in terms of the generator of the relevant group. So if $x$ is the boundary, then it is $1$ in terms of the generator for $H_1(A\cap B)$ (which is $x$ itself) but it is $2$ in terms of the generator of $H_1(A)$, which is the core circle (likewise for $B$). I misinterpreted $x \mapsto (x,-x)$ as being a map defined using generators (so that each $x$ represents a generator). Thanks! – Anon Jan 08 '23 at 20:42