Let $ M $ and $ \mathbb{T}^{2} $ denote the Möbius strip and the torus respectively. Suppose that we attach $ M $ to $ \mathbb{T}^{2} $ by wrapping the boundary circle $ C $ of $ M $ around the first circle of the torus $ k $ times, i.e., we have a continuous attaching map $ f: C \to \mathbb{S}^{1} \times \{ 1 \} \subseteq \mathbb{T}^{2} $ with winding number $ k $. Then what would be the first homology group of the resulting adjunction space $ X = M \sqcup_{f} \mathbb{T}^{2} $?
Set theoretically, we have
$$
X = (M \times \{ 0 \}) \cup (\mathbb{T}^{2} \times \{ 1 \}) \Big/ \sim,
$$
where $ \sim $ is the equivalence relation defined by $ (x,0) \sim (f(x),1) $ for all $ x \in M $. Next, consider the following two subspaces of $ X $:
$$
A = [M \times \{ 0 \}]_{\sim} \quad \text{and} \quad
B = [\mathbb{T}^{2} \times \{ 1 \}]_{\sim}.
$$
Their intersection is $ [C \times \{ 0 \}]_{\sim} = [(\mathbb{S}^{1} \times \{ 1 \}) \times \{ 1 \}]_{\sim} $. Pick an open neighborhood $ U $ of $ A $ and an open neighborhood $ V $ of $ B $ such that $ U $ and $ V $ deformation retract onto $ A $ and $ B $ respectively. Then $ U \cap V $ deformation retracts onto $ A \cap B = [C \times \{ 0 \}]_{\sim} $. As $ \{ U,V \} $ is an open cover of $ X $, we can apply the Mayer-Vietoris sequence to it.
My problem is figuring out what the mapping
$$
\require{AMScd}
\begin{CD}
{H_{1}}(U \cap V) @>{i_{*} \oplus j_{*}}>> {H_{1}}(U) \oplus {H_{1}}(V)
\end{CD}
$$
looks like. Deformation retraction reduces this problem to analyzing the mapping
$$
\begin{CD}
{H_{1}}([C \times \{ 0 \}]_{\sim}) @>{i_{*} \oplus j_{*}}>>
{H_{1}}([M \times \{ 0 \}]_{\sim}) \oplus {H_{1}}([\mathbb{T}^{2} \times \{ 1 \}]_{\sim}).
\end{CD}
$$
My questions are:
-
Is $ [C \times \{ 0 \}]_{\sim} $ homeomorphic to $ \mathbb{S}^{1} $? This is not very straightforward because $ \sim $ collapses sets of $ k $ points in $ C \times \{ 0 \} $ to singletons. In the case $ k = 2 $, I know that the answer is ‘yes’ because $ \sim $ restricted to $ C \times \{ 0 \} $ behaves as if it were identifying antipodal points in $ \mathbb{S}^{1} $, and it is well-known that identifying antipodal points in $ \mathbb{S}^{1} $ produces $ \mathbb{RP}^{1} $, which is homeomorphic to $ \mathbb{S}^{1} $.
-
Suppose that the answer to the above question is ‘yes’ for $ k $ in general. Then how does the homology class of a fundamental cycle $ \gamma $ in $ [C \times \{ 0 \}]_{\sim} $ map to $ {H_{1}}([M \times \{ 0 \}]_{\sim}) $ under $ i_{*} $? I believe that the answer to this is
$$
(i_{*})([\gamma]) = \pm 2 \cdot [\text{Generating core loop of $ M $}],
$$
where the sign depends on which orientation we are using. -
Similarly, how does $ [\gamma] $ map to $ {H_{1}}([\mathbb{T}^{2} \times \{ 1 \}]_{\sim}) $ under $ j_{*} $?
Best Answer
EDIT: There was some mistakes here, I meant to write $ [(\mathbb{S}^{1} \times \{ 1 \}) \times \{ 1 \}] $. Here is the right answer for $ C $: since $ f $ is a degree $ k$ map, we can homotope to the standard map $ f_k: e^{i \theta} \rightarrow e^{ i k \theta} $. This gives a homotopy $ S^1/f $ to $ S^1/f_k = S^1 $.
The second question is crucial. If you go around the circle in $ S^1 \subset \mathbb{T}^2 $ once, then the attaching map will take you around $ S^1 = \partial M $, $ k $ times over. When you go to homology, this goes to $ 2 k $, so overall $ [ \gamma ] \rightarrow 2k [ \gamma ] $. It will only go around the $ \mathbb{T}^2 $ once, so the map is just $ [ \gamma ] \rightarrow [ \gamma ] $.
I hope this will help!