The First Homology Group Obtained by Attaching a Möbius Strip to a Torus in a Certain Way.

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 Caleb Fillmore May 14 '14 at 2:07
AKA, for those about to help, we salute you! – user99680 May 14 '14 at 2:10

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 $[ \gamma ] \rightarrow [ \gamma ]$.
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 Caleb Fillmore May 14 '14 at 4:58
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 '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 Caleb Fillmore May 15 '14 at 0:20