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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_{*} $?

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I forgot to say: A word of thanks to those who are willing to help! –  Berrick Fillmore May 14 at 2:07
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AKA, for those about to help, we salute you! –  user99680 May 14 at 2:10

1 Answer 1

up vote 1 down vote accepted

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!

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Thanks for your response! Actually, I think that points in $ C $ are identified. Each $ p \in C $ is identified with $ k - 1 $ other points in $ C $ because $ \#({f^{\leftarrow}}[f[\{ p \}]]) = k $. I initially thought that points in $ C $ remained distinct, but I abandoned this point of view later on. –  Berrick Fillmore May 14 at 4:58
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Ah I made a typo. I think it will be tricky to show that there is a homeomorphism on $ C $. It's not true that each $ p \in C $ is identified with $ k-1 $ other points - you could imagine $ f $ oscillating arbitrary number of times, but still having degree $ k $. I think this is why it's easier to work in with $ [(\mathbb{S}^{1} \times \{ 1 \}) \times \{ 1 \}]_{\sim} $, where really the equivalence $ \sim $ does not identify points. –  user149943 May 14 at 6:43
    
Hence, is it right to say that the mapping $ {H_{1}}(U \cap V) \to {H_{1}}(U) \oplus {H_{1}}(V) $ is given by $ 1 \mapsto (2 k,(1,0)) $ for a suitable basis of $ {H_{1}}(U) \oplus {H_{1}}(V) $? –  Berrick Fillmore May 15 at 0:20

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