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?
Best Answer
$A \cap B$ 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.