(This is exercise 2.2.28 from Hatcher) Consider the space obtained from a torus $T^2$, by attaching a Mobius band $M$ via a homeomorphism from the boundary circle of the Mobius band to the circle $S^1\times \{x_0\} \subset T^2$.

We set $A=T^2$, $B=M$, $X:=A\cup B $, $A\cap B\simeq S^1$, so the (reduced) Mayer-Vietoris sequence yields

$$ 0\to \tilde H_2(T^2)\oplus\tilde H_2(M)\to \tilde H_2(X)\to \tilde H_1(S^1)\to \tilde H_1(T^2)\oplus \tilde H_1(M)\to \tilde H_1(X) \to 0$$ and replacing the homology groups we already know we have $$0\to \mathbb Z\overset{\alpha}{\to} \tilde H_2(X) \overset{\beta}{\to} \mathbb Z \overset{\gamma}{\to} \mathbb Z^2\oplus \mathbb Z\overset{\delta}{\to} \tilde H_1(X)\to 0$$

Now, I have to figure out what the maps $\alpha,\beta,\gamma,\delta$ do, and this is where I'm stuck. I'd appreciate a detailed explanation.


1 Answer 1


The map $\gamma$ is $(i_*, j_*)$ where $i : S^1 \hookrightarrow T^2$, $j : S^1\hookrightarrow M$. Since $i_*$ maps the generator of $H_1(S^1)$ to a generator of $H_1(T^2)$, and $j_*$ maps the generator of $S^1$ to twice the generator of $H_1(M)$, we find that $\gamma(1) = ((1, 0), 2)$. From the exactness of the sequence, $$ H_1(X) \cong H_1(T^2) \oplus H_1(M) / \operatorname{im} \gamma \cong \langle x, y, z \mid x + 2z\rangle_{ab}. $$

Since $x = -2z$ in the presentation, we can remove $x$ to get $$ H_1(X) \cong \langle y, z \rangle_{ab} \cong \Bbb Z \oplus \Bbb Z. $$

Since $\operatorname{im} \beta = \ker \gamma = 0$, it follows that $H_2(X) = \Bbb Z$.

It's worth noting that $A$, $B$ must be open when applying MVS. $T^2$ and $M$ are not open in $X$, but this is easy to fix: Just take an open neighborhood that deformation retracts onto each subspace.

  • $\begingroup$ Can you also explain what the map $\delta$ does to the generators of $\tilde H_1(T^2)\oplus \tilde H_1(M)$? $\endgroup$
    – Dimitris
    Commented May 26, 2014 at 23:19
  • $\begingroup$ Also, the second part of the question asks what happens if we replace $T^2$ by $\mathbb R P^2$, so we attach $M$ along the standard $\mathbb RP^1\subset \mathbb RP^2$. In this case what would $(i_*,j_*)$ do? I guess $i_*$ would send the generator of $H_1(S^1)$ to either $1$ or $0$ depending on how many times $S^1$ wraps around in $\mathbb RP^2$. $\endgroup$
    – Dimitris
    Commented May 26, 2014 at 23:47
  • 1
    $\begingroup$ @DimitriosNtalampekos $\delta$ is slightly more difficult to visualize. The trick with MVS is to pick the map that is easier to work with, and compute the rest via exactness. Anyway, $\delta$ maps $z$ to a generator (cycle in $M$), $y$ to the other generator (cycle in $T^2$, not the one identified with the boundary of $M$), and $x$ to twice the image of $z$, but in the opposite direction. $\endgroup$ Commented May 27, 2014 at 8:56
  • $\begingroup$ @DimitriosNtalampekos As for your other question, I think you got it. Write the long exact sequence, and compute the homology groups as I did. $\endgroup$ Commented May 27, 2014 at 8:59
  • 1
    $\begingroup$ @sifsa Those notes essentially claim that $\mathbb Z \oplus \mathbb Z / \langle 1, 2\rangle \cong \mathbb Z \oplus \mathbb Z_2$. This is wrong. $\endgroup$ Commented May 14, 2016 at 18:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .