Compute homology groups of space $\Bbb RP^2$ attached with Mobius band using Mayer Vietories

Question

(This is exercise 2.2.28 from Hatcher) Consider the space $X$ (say) obtained from a $\Bbb RP^2$, by attaching a Mobius band $M$ via a homeomorphism from the boundary circle of the Mobius band to the circle $\Bbb RP^1 \subset \Bbb RP^2$. Compute the homology of the space

We set $A=N_{\epsilon}(\Bbb RP^2)$, $B=N_{\epsilon}(M)$, $X:=A^o\cup B^o $, s.t $A$, $B$ and $A\cap B$ deformation retracts to $\Bbb RP^2$, $M$ and $ S^1$, so the (reduced) Mayer-Vietoris sequence yields

$$ \tilde H_2(S^1)\to \tilde H_2(\Bbb RP^2)\oplus\tilde H_2(M)\to \tilde H_2(X)\to \tilde H_1(S^1)\to \tilde H_1(\Bbb RP^2)\oplus \tilde H_1(M)\to \tilde H_1(X) \to 0$$ and replacing the homology groups we already know we have $$0\to 0 \to \tilde H_2(X) \overset{\partial}{\to} \mathbb Z \overset{\phi}{\to} \mathbb Z_2\oplus \mathbb Z\overset{\psi}{\to} \tilde H_1(X)\to 0$$

Now, I have to figure out what the maps $\partial,\phi,\psi$ do, and this is where I'm stuck. I'd appreciate a detailed explanation.

What I am thinking that $\partial$ is injective and $\tilde H_1(X)= (\mathbb Z_2\oplus \mathbb Z)/ Im \phi$. Again $\phi=(j_*,i_*)$ where $i: S^1 \hookrightarrow M , j: S^1 \hookrightarrow \Bbb RP^2$. Then $i_*(1)=2$, but little bit confused about $j_*$ is $j_*(1)=[0]$ because in calculating homology of $\Bbb RP^2$ the map from $e^1 \to e^0$ is zero map again I am thinking that the map might be $1 \mapsto [1]$. Now for $j_*(1)=[0]$.

$\tilde H_1(X)= (\mathbb Z_2\oplus \mathbb Z)/ Im \phi= (\mathbb Z_2\oplus \mathbb Z)/<(2,[0])>=\Bbb Z_2 \oplus \Bbb Z_2$. Then $ker \phi=0=Im \partial$. So, $\tilde H_2(X)=0$. Can anyone please tell me whether I am right or wrong or how to rectify it, moreover what is $ H_0(X)$?

Answer 1

I can see it has been some years since this question was asked, but since it has no answers, let me have a go.

Looking at $0 \mapsto H_{2}(X) \mapsto \mathbb{Z} \mapsto \mathbb{Z}_{2} \oplus \mathbb{Z} \mapsto H_{1}(X) \mapsto 0$. Let's find $\phi$ first. $\phi$ is induced by the inclusion of $S^{1} \cong \mathbb{R}P^{1}$ into $\mathbb{R}P^{2}$ and its inclusion in $M$. Seeing that $\mathbb{R}P^{1}$ wraps around the mobius band twice we see that $\mathbb{Z} \mapsto\mathbb{Z}/2\oplus \mathbb{Z}$ is $1 \mapsto ([1],2)$. This is injective so $H_{2}=0$. $H_{1}(X)=\frac{\mathbb{Z}/2\oplus\mathbb{Z}}{([1],2)}\cong \mathbb{Z}/4.$ $H_{0}(X)\cong\mathbb{Z}$ being connected.

Answer 2

We have that $j$ takes $S^1$ onto $S^1\cong\mathbb{R}P^1\subset\mathbb{R}P^2$, a cycle that represents the generator of $H_1(\mathbb{R}P^2)$, so $j_*(1)=[1]$ and $\text{im}(\phi)=<([1],2)>.$ The kernel of $\phi$ is still zero, so $H_2(X)=0$.

As commented, $H_0(X)=\mathbb{Z}$ since $X$ is path connected.

We will argue $H_1(X)\cong \mathbb{Z}/4\mathbb{Z}$. We have the following partition $$\{([k],2k): k\in\mathbb{Z}\}\sqcup\{([k+1],2k): k\in\mathbb{Z}\}\sqcup\{([k],2k+1): k\in\mathbb{Z}\}\sqcup\{([k+1],2k+1): k\in\mathbb{Z}\}\\ = \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}.$$

Under the quotient map $\pi: \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z} \to (\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z})/<([1],2)> \cong(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}) / \text{im}(\phi)$,

$$\pi([k],2k)=[([0],0)] $$ $$\pi([k+1],2k)=[([1],0)] $$ $$\pi([k],2k+1)=[([0],1])] $$ $$\pi([k+1],2k+1)=[([1],1)]. $$

With $H_1(X)\cong (\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z})\ /\ \text{im}(\phi)$, the quotient map shows $|H_1(X)|=4$. Since $2[([0],1)]=[([0],2)]$ and $([0],2)=([k+1],2k)$ for $k=1$, we have $2[([0],1)]=[([1],0)]\neq[([0],0)]$ implying $H_1(X)\neq \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}.$ Thus, $H_1(X)\cong \mathbb{Z}/4\mathbb{Z}$, the only other group of order 4.

Answer 3

This is also not too hard using cellular homology. To set this up correctly we first need to understand how to cellulate the Mobius strip:

I don't claim this is the only way but hey, it's convenient. We use two $0$-cells $x,y$; we use $3$ $1$-cells $\alpha,\ell_1,\ell_2$ which run as in the picture (so that, modulo signs, $(-\ell_1)\cdot\ell_2$ describes the boundary circle). We attach one $2$-cell $\sigma$ via regarding the boundary $S^1$ as the loop $\ell_1\cdot\alpha\cdot\ell_2\cdot\alpha$. Take a moment to convince yourself this winds correctly to give the Mobius strip.

We can then cellulate our space $X$ by slightly modifying the usual description of $\Bbb RP^2$. We keep $\ell_1,\ell_2,x,y$ as a cellulation of $S^1$; then we know attaching a $2$-cell via $S^1\twoheadrightarrow S^1=\Bbb RP^1$ will give a homeomorph of $\Bbb RP^2$, so we do this and call the attached $2$-cell $\sigma'$. Since the quotient map there is just $z\mapsto z^2$, the double-winding, we could describe this attaching via the doubled loop $(-\ell_1)\cdot\ell_2\cdot(-\ell_1)\cdot\ell_2$. We can easily attach a Mobius strip to $\Bbb RP^1\subset\Bbb RP^2$ here, along the boundary circle as required; adjoin another $1$-cell which we call $\alpha$, as before, which runs $y\to x$, and attach a second $2$-cell $\sigma$ which, as before, has boundary loop $\ell_1\cdot\alpha\cdot\ell_2\cdot\alpha$. $$X=\overset{\sigma=\text{ Mobius strip},\,\alpha}{\overbrace{e_2\cup e_1}}\cup\overset{\Bbb RP^2}{\overbrace{e_2\cup\underset{S^1=\text{ boundary circle}}{\underbrace{e_1\cup e_1\cup e_0\cup e_0}}}}$$

The homology of the space is then computable by the equivalent cellular complex: $$0\to\Bbb Z^2\to\Bbb Z^3\to\Bbb Z^2\to0$$The first boundary $\partial_1:\Bbb Z^3\to\Bbb Z^2$ is the simplicial boundary map, taking $\alpha\mapsto x-y;\ell_1,\ell_2\mapsto y-x$. From this, the obvious fact $H_0(X)\cong\Bbb Z$ is recovered.

Identifying the first, second and third coordinates of $\Bbb Z^3$ with $\alpha,\ell_1,\ell_2$ respectively, the kernel of $\partial_1$ is $\{(u+v,u,v):u,v\in\Bbb Z\}$.

To calculate $\partial_2$ using the degree formula is easy; the doubly-wound $\sigma'$ has incidence $0$ around $\alpha$, $-2$ around $\ell_1$ and $2$ around $\ell_2$ (up to a factor of $(-1)$, as usual). $\sigma$ similarly has incidence $2,1,1$ so that: $$\partial_2(s,t)=(2t,t-2s,t+2s)$$The kernel is trivial, so $H_n(X)\cong0$ for all $n\ge2$.

We finally have: $$H_1(X)\cong\frac{\{(v+w,v,w)\}}{\{(2t,t-2s,t+2s)\}}\cong\Bbb Z_4$$The idea is that an element of the numerator lies in the denominator if and only if $w-v=(t+2s)-(t-2s)=4s$ is divisible by $4$; indeed the map $\ker\partial_1\to\Bbb Z_4$, $(u,v,w)\mapsto[w-v]$ witnesses this quotient.

It is interesting to note that if one first attaches the Mobius strip around the equator of $S^2$ and then applies the quotient $S^2\to\Bbb RP^2$, the first homology is $\Bbb Z_2\oplus\Bbb Z_2$ instead.