Consider the surface $M_g$ of genus $g$, embedded in $\Bbb{R}^3$ in the standard way. It bounds some compact region $R$. Two copies of $R$ are glued together by the identity map between their boundary surfaces, which forms a closed 3-manifold $X$. I am asked to compute the homology groups of $X$.

Now the computation of this for $n = 3$ of $H_n(X)$ is straightforward from Mayer - Vietoris. However now for $n=2$, I run into trouble: I am looking at the following part of the LES from Mayer - Vietoris. Put $A$ = one copy of $R$, $B$ = the other copy, their intersection $A \cap B = M_g$. $i,j$ are the inclusion maps of $A \cap B$ into $A$ and $B$ respectively. The lower end of Mayer - Vietoris looks like

$$0 \rightarrow \tilde{H}_2(X) \rightarrow \tilde{H}_1(A \cap B) \stackrel{(i_\ast,j_\ast)}{\longrightarrow} \tilde{H}_1(A) \oplus \tilde{H}_1(B) \rightarrow \tilde{H_1}(C) \rightarrow 0 \rightarrow 0 \rightarrow 0 \rightarrow 0.$$

Now I know that $\tilde{H_1}(A \cap B) = \Bbb{Z}^{2g}$ and the same for $ \tilde{H}_1(A) \oplus \tilde{H_1}(B)$, but this does not imply that $(i_\ast,j_\ast)$ is an isomorphism. The first fact on the homology of $A \cap B$ comes from the CW - structure of $M_g$ having $2g$ one cells, the second from the fact that $A$ and $B$ respectively can be thought of the wedge sum of $g$ tori, which is homotopy equivalent to a wedge of $g$ circles.

Now the complete the problem I need to know the kernel of the map $\Phi$. To do this, I need to know what are

1. Generators for the homology of $M_g$.

2. Generators for the homology of $A$ and $B$.

How do I go about finding these? I would say my main problem in general is making connections between algebraic things and generators for homologies that comes from topology.

  • $\begingroup$ Perhaps I'm wrong (my algebraic topology is not strong) but doesn't Mayer-Vietoris require that the interiors cover the space, while here they don't? $\endgroup$
    – only
    Commented Oct 21, 2012 at 13:55
  • 3
    $\begingroup$ @only Well there are $\epsilon$ fattenings of $A$ and $B$ that deformation retract back to them so it does not matter. $\endgroup$
    – user38268
    Commented Oct 21, 2012 at 13:55
  • $\begingroup$ I see. On the other hand, you prove that $\tilde{H}_1(A) \oplus \tilde{H}_1(B)$ is isomorphic to $\tilde{H_1}(X)$ (I assume your $C$ is a typo for $X.$) Doesn't this prove that $\Phi$ is the zero map? $\endgroup$
    – only
    Commented Oct 21, 2012 at 14:01
  • $\begingroup$ @only Sorry, I meant to write $A \cap B = M_g$. $\endgroup$
    – user38268
    Commented Oct 21, 2012 at 14:06
  • $\begingroup$ @only I have edited my answer. $\endgroup$
    – user38268
    Commented Oct 21, 2012 at 14:07

1 Answer 1


I have to think about it and don't have the time, but I would say the following:

  • I would start with $g =1$. That's the standard trick. :-)
  • In this case, I would remember that your morphism $\Phi$ is what I called in my answer $(i_*, j_*)$, the one induced by the inclusions $A \leftarrow A \cap B \rightarrow B$.
  • Then, we can take the generators for the $H_1$ of $A \cap B = M_1 = \mathbb{T}^2$ to be two $S^1$: one "meridian" and one "parallel". Take the equatorial inner one for the later, for instance.
  • What happens with these generators inside of $A = B= R$? (That is, once we apply $i_*$ and $j_*$ to them?) Well, the first one, the meridian, goes to zero and the second one, the parallel, survives as himself. Doesn't them?

So, you've got your morphism $\Phi = (i_*, j_*)$ and you can pursue your computations, I think.

EDIT. You've probably already guessed it, but for $g=2$ , you have four generators: two meridians and two parallels too. Meridians go to zero through $(i_*, j_*)$. Parallels remain the same. And so on: in general, you'll have $2g$ generators...

  • $\begingroup$ Thanks for your answer. However why is it that the meridian goes to zero while the parallel survives itself? It seems to me that applying $(i_\ast,j_\ast)$ to both of them do nothing, this is where I'm confused. For example, $(i_\ast,j_\ast)[\alpha] = ([\alpha],[-\alpha])$ calling $\alpha$ the generator corresponding to the meridian. Now why should that be zero? $\endgroup$
    – user38268
    Commented Oct 21, 2012 at 22:41
  • $\begingroup$ Because the meridian, inside the full torus $R$, contracts to a point (or bounds a whole disc there, if you prefer; anyway, it's a boundary there). $\endgroup$ Commented Oct 22, 2012 at 2:52

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