----- > [!proposition] Proposition. ([[Homology of the Klein bottle]]) > The [[singular homology]] of the [[Klein bottle]] $K$ is $H_{i}(K)=\begin{cases} 0 & i \geq 2 \\ \mathbb{Z} \oplus \frac{\mathbb{Z}}{2\mathbb{Z}} & i=1 \\ \mathbb{Z} & i=0. \end{cases}$ ^proposition - page 22 of notes - using [[Mayer-Vietoris theorem]] - page 43 of notes - using [[cellular homology]] **Proof via Mayer-Vietoris.** Cover $K$ with opens $A,B$ as depicted: ![[Pasted image 20250603090729.png|500]] Extract $A,B$, and the intersection $A \cap B$. $B,A$ are each [[homotopy equivalent|homotopy equivalent]] to circles $b,a$ respectively; $A \cap B$ is homotopy equivalent to two disjoint circles $s, t$. The circles are labeled in green. ![[Pasted image 20250603092221.png]] Now we have to think about what the circles $s,t$ do when we include them back into $B$ and back into $A$. This looks as follows: ![[Pasted image 20250603093212.png]] Now, $s$ and $t$ are 1-cycles generating the first homology of their respective circles, so we write $H_{1}(A \cap B)\cong H_{1}(\mathbb{S}^{1} \sqcup \mathbb{S}^{1})\cong \mathbb{Z}\langle s \rangle \oplus \mathbb{Z} \langle t \rangle .$ Running [[Mayer-Vietoris theorem|Mayer-Vietoris]], and labeling what we know so far: $\begin{align} 0\to H_{2}(K) &\xrightarrow{\partial_{MV}} \overbrace{ H_{1}(A \cap B) }^{ \mathbb{Z}\langle s \rangle \oplus \mathbb{Z}\langle t \rangle } \xrightarrow{(\iota_{A_{*}}, \iota_{B_{*}})} \overbrace{ H_{1}(A) }^{ \mathbb{Z}\langle a \rangle } \oplus \overbrace{ H_{1}(B) }^{ \mathbb{Z}\langle b \rangle }\xrightarrow{j_{A_{*}}-j_{B_{*}}} H_{1}(K) \\ &\xrightarrow{\partial_{MV}} \underbrace{ H_{0}(A \cap B) }_{ \mathbb{Z} \langle x \rangle \oplus \mathbb{Z} \langle y \rangle }\xrightarrow{(\iota_{A_{*}, }\iota_{B_{*}})} \underbrace{ H_{0}(A) }_{ \mathbb{Z} } \oplus \underbrace{ H_{0}(B) }_{ \mathbb{Z} } \xrightarrow{ j_{A_{*}}-j_{B_{*}}} H_{0}(K) \\ & \to 0. \end{align}$ First note $H_{i}$ manifestly vanishes for $i \geq 2.$ And $H_{0}=\mathbb{Z}$ because [[connected]]. So we just need to find $H_{2}$ and $H_{1}$. **Finding $H_{2}$.** To show $H_{2}=0$, it suffices to show $\partial_{MV}^{1}=0$. By exactness, this is equivalent to showing $\operatorname{ker }(\iota_{A_{*}}, \iota_{B_{*}})=0$. Indeed, we can understand the map on first homology $(\iota_{A_{*}}, \iota_{B_{*}}):\mathbb{Z}\langle s \rangle \oplus \mathbb{Z}\langle t \rangle \to \mathbb{Z}\langle a \rangle \oplus \mathbb{Z}\langle t \rangle$, by understanding how $s,t$ include into $A,B$. $s$ includes into $A$ as just $a$, while $t$ includes into $a$ with a twist, i.e., as $-a$. That is, $\iota_{A_{*}}(s)=a$ and $\iota_{A_{*}}(t)=-a$. $s$ and $t$ both include into $B$ as $b$, that is, $\iota_{B_{*}}(s)=b=\iota_{B_{*}}(t)$. So the map $(\iota_{A_{*}}, \iota_{B_{8}})$ is represented by the matrix $\begin{bmatrix}1 & -1 \\1 & 1\end{bmatrix}$ (or $\begin{bmatrix}1& 1 \\-1 & 1\end{bmatrix}$?) which is invertible; hence the map is in particular injective. So $\operatorname{im }\partial_{MV}^{1}=0$, hence $H_{2}(K)=0$. **Finding $H_{1}$.** Recall from [[exact sequence wrangling]] that if we have an [[exact sequence]] where we know $A,B$, $f$, $D,E$, $g$: $A \xrightarrow{f} B \to C \to D \xrightarrow{g} E$ then the sequence $0 \to \text{coker }f \to C \to \text{ker }g \to 0$is [[short exact sequence|short exact]]. To spoil: here, the sequence in question will be $\mathbb{Z} \langle s, t \rangle \xrightarrow{f=\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}} \mathbb{Z} \langle a \rangle \oplus \mathbb{Z} \langle b \rangle \to H_{1}(K) \to \mathbb{Z} \langle x, y \rangle \xrightarrow{g=\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}} \mathbb{Z} \oplus \mathbb{Z} $ and [[exact sequence wrangling]] will give a [[short exact sequence]] $0 \to \underbrace{ \frac{\mathbb{Z}}{2\mathbb{Z}} }_{ \operatorname{coker }f } \to H_{1}(K) \to \underbrace{ \mathbb{Z} }_{ \operatorname{ker }g } \to 0$ which [[split short exact sequence|splits]] because $\mathbb{Z}$ is [[free abelian group|free]] + [[the splitting lemma]]. --- So let us show this. First look at the map on zeroth homology $g=(\iota_{A_{*}}, \iota_{B_{*}})$. We just look at inclusions of path components and see the matrix is $g=\begin{bmatrix}1 & 1 \\1 & 1\end{bmatrix}$. $\operatorname{ker }g$ is $\text{span}(\begin{bmatrix}1 \\ -1\end{bmatrix}) \cong \mathbb{Z}\langle x-y \rangle$. Next note the image of $f=\begin{bmatrix}1 & -1 \\1 & 1\end{bmatrix}$ is generated by $\{ a+b, -a+b \}$; equivalently by $\{ -a+b, 2b \}$ (adding the two generators). Thus $\operatorname{coker }f=\frac{\overbrace{ H_{1}(A) \oplus H_{1}(B) }^{ \mathbb{Z}\langle a , b\rangle }}{\underbrace{ \operatorname{im }(\iota_{A_{*}}, \iota_{B_{*}}) }_{ \mathbb{Z} \langle -a+b, 2b \rangle }}=\frac{\mathbb{Z}}{2\mathbb{Z}} \langle b \rangle.$ We are done. > [!proof]- Proof. ([[Homology of the Klein bottle]]) > ~ ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```