(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)$?

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    $\begingroup$ No the question is different look the question there it is $\Bbb RP^2$ instead of $T^2$. $\endgroup$
    – Ri-Li
    Mar 15, 2015 at 7:23
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    $\begingroup$ No the question is different look the question there it is $\Bbb RP^2$ instead of $T^2$. In the comment also the solution is not clearly written. So, anyone can see my calculation also.. and check the proof. $\endgroup$
    – Ri-Li
    Mar 15, 2015 at 7:37
  • $\begingroup$ What you have to do is to describe explicitely the map $\phi$. $\endgroup$ Mar 15, 2015 at 9:52
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    $\begingroup$ As for $H_0(X)$: you should know by now that $H_0(X)$ is a free abelian group on as many generators as there are path components in $X$. $\endgroup$ Mar 15, 2015 at 9:53

3 Answers 3


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.

  • $\begingroup$ I think it should be $H_1(X) \cong \mathbb{Z}/4$. You've written $([1],1)$ instead of $([1],2)$ on the last line, which changes the answer. $\endgroup$ May 3, 2021 at 10:24
  • $\begingroup$ I haven't touched algebraic topology in a while, but 2 makes sense to me. The first homology group of the mobius band is generated by the equator of the mobius band. Now, if you look at the boundary of the mobius band (i.e. what the inclusion of $\mathbb{R}P^{1}$ into $M$ gives you), then this is homologous to twice the equator (the boundary wraps around twice). Hence, the second slot should be a 2 and not a 1. $\endgroup$ May 4, 2021 at 9:41
  • $\begingroup$ Yes, I agree. You've written that the map is $1 \mapsto ([1],2)$ on the penultimate line and then that its image is $([1], 1)$ on the final line. I'm saying I think the image on the final line should be $([1], 2)$. $\endgroup$ May 4, 2021 at 9:43
  • $\begingroup$ oh I see - my bad - I'll edit my answer. $\endgroup$ May 4, 2021 at 9:49
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    $\begingroup$ I still think the answer should be $\mathbb{Z}/4$ though, since the element $([1], 1)$ has order four in $(\mathbb{Z}/2\oplus \mathbb{Z})/([1],2)$, so the group can't be $\mathbb{Z}/2\oplus \mathbb{Z}/2$. $\endgroup$ May 4, 2021 at 10:38

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.


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:

Annotated Mobius strip with a CW structure

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.


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