If I choose two open sets $A$ and $B$ as depicted on Wikipedia here:

Mayer-Vietoris for the Torus

then I have an isomorphism between $H_n(A \cap B)$ and $H_n(A) \oplus H_n(B)$ because the two tubes in $A \cap B$ are disjoint.

OK, so far so good. Then I write down the Mayer-Vietoris sequence and try to compute $H_n(T^2)$ but it seems to me I don't have enough information to do so. Can you confirm this?

I then computed the reduced MVS instead and used that to compute $H_n(T^2)$. What I would like to know is if this is the only way or if there is a way to do it with MVS directly, without the reduced MVS.

Many thanks for your help, I appreciate it.

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    $\begingroup$ What do you mean, an isomorphism between $H_n(A\cap B)$ and $H_n(A)\oplus H_n(B)$? They are isomorphic as groups, but the inclusions don't induce isomorphisms! For example, the map from $H_1(A\cap B)$ to $H_1(A)\oplus H_1(B)$ has kernel $(1,-1)$, where $H_1(A\cap B)\cong \mathbb{Z}\oplus\mathbb{Z}$. $\endgroup$
    – user641
    Aug 18, 2011 at 18:12

2 Answers 2


Maybe I can add some details to Dylan's answer.

First of all, with the open cover you have chosen, you have

$$ A, B \quad \cong \quad \text{cylinder} \quad \simeq \quad S^1 $$


$$ A \cap B \quad \cong \quad \text{disjoint union of two cylinders} \quad \simeq \quad S^1 \sqcup S^1 \ . $$


$$ H_n(A) = H_n(B) = \begin{cases} \mathbb{Z} & \text{if}\ n=0,1 \\ 0 & \text{otherwise} \end{cases} $$


$$ H_n(A\cap B) = \begin{cases} \mathbb{Z}\oplus \mathbb{Z} & \text{if}\ n=0,1 \\ 0 & \text{otherwise} \end{cases} $$

Hence, in particular, for $n\geq 2$, $H_n(A) = H_n(B) = H_n(A\cap B) = 0$. Thus, from the MVS,

$$ \dots \longrightarrow H_n(A)\oplus H_n(B) \longrightarrow H_n(\mathbb{T}^2) \longrightarrow H_{n-1}(A \cap B) \longrightarrow \dots $$

which, for $n > 2$, is just

$$ \dots \longrightarrow 0 \longrightarrow H_n(\mathbb{T}^2) \longrightarrow 0 \longrightarrow \dots $$

follows that

$$ H_n(\mathbb{T}^2) = 0, \quad \text{for} \quad n> 2 . $$

So, again we can focus on what happens for $n=0,1,2$. For $n=0$ there is no big deal, because, since $\mathbb{T}^2$ is connected,

$$ H_0 (\mathbb{T}^2) = \mathbb{Z} \ . $$

For $n= 2$, we look at this piece of the MVS:

$$ \dots \longrightarrow H_2(A)\oplus H_2(B) \longrightarrow H_2(\mathbb{T}^2) \stackrel{\partial}{\longrightarrow} H_1(A\cap B) \stackrel{(i_*, j_*)}{\longrightarrow} H_1(A) \oplus H_1(B) \longrightarrow \dots \ , $$

of which we know all the groups, except the torus' one:

$$ \dots \longrightarrow 0 \longrightarrow H_2(\mathbb{T}^2) \stackrel{\partial}{\longrightarrow} \mathbb{Z}\oplus\mathbb{Z} \stackrel{(i_*, j_*)}{\longrightarrow} \mathbb{Z}\oplus\mathbb{Z} \longrightarrow \dots $$

Now, as Steve D. pointed you out, despite the two last groups being isomorphic, it doesn't mean that the morphism $(i_*, j_*)$ between them in the MVS, induced by the inclusions $i: A\cap B \longrightarrow A$ and $j: A\cap B \longrightarrow B$, is an isomorphism. And actually it is not. (If it was, then $H_2(\mathbb{T}^2) $ would be zero.)

At this stage, you need to compute who this $(i_*, j_*)$ is. For this, you choose $1$-cicles generating the homologies of $A, B $ and $A\cap B$ as follows: for each cylinder of the intersection $A\cap B$, take an equatorial circumference. Name their homology classes $\alpha$ and $\beta$. So, actually, those $\mathbb{Z}$ in the piece of MVS depicted above are the free abelian groups generated by $\alpha$ and $\beta$:

$$ (i_*, j_*) : \mathbb{Z}\langle \alpha \rangle \oplus \mathbb{Z}\langle \beta \rangle \longrightarrow \mathbb{Z}\langle \alpha \rangle \oplus \mathbb{Z}\langle \beta \rangle $$

And now we compute:

$$ (i_* , j_*) (\alpha , 0) = (i_* , j_*) (0, \beta ) = (\alpha , \beta) $$

since $\alpha = \beta$ in $H_1(A)$ and $H_1(B)$.

Hence, in terms of these basis, our morphism $(i_* , j_*)$ can be represented by the matrix

$$ \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} : \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} $$


$$ H_2(\mathbb{T}^2 ) = \mathrm{im}\ \partial = \mathrm{ker}\ \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \mathbb{Z}\langle \alpha - \beta \rangle = \mathbb{Z} $$

As for $H_1(\mathbb{T}^2)$, let's focus on the following piece of the MVS:

$$ \dots \longrightarrow H_1(A\cap B) \stackrel{(i_*, j_*)}{\longrightarrow} H_1(A) \oplus H_1(B) \stackrel{k_* - l_*}{\longrightarrow} H_1(\mathbb{T}^2) \stackrel{\partial}{\longrightarrow} H_0 (A\cap B) \stackrel{(i_* , j_*)}{\longrightarrow} H_0(A) \oplus H_0(B) \longrightarrow \dots \ , $$

where $k_* - l_*$ is the morphism induced by the inclusions $k: A \longrightarrow \mathbb{T}^2$ and $l : B \longrightarrow \mathbb{T}^2$.

Again, we know all the groups except $H_1(\mathbb{T}^2)$:

$$ \dots \longrightarrow \mathbb{Z}\oplus\mathbb{Z} \stackrel{(i_*, j_*)}{\longrightarrow} \mathbb{Z}\oplus\mathbb{Z} \stackrel{k_* - l_*}{\longrightarrow} H_1(\mathbb{T}^2) \stackrel{\partial}{\longrightarrow} \mathbb{Z}\oplus\mathbb{Z} \stackrel{(i_* , j_*)}{\longrightarrow}\mathbb{Z}\oplus\mathbb{Z} \longrightarrow \dots $$

And again, we need some knowledge of the morphisms involved. I claim that, taking as generators for $H_0(A)$ and $ H_0(B)$ two points $p,q$, one in each component of $A\cap B$, you have also generators for $H_0(A\cap B)$ and, with these generators and similar computations that the ones we have already done, the second morphism $(i_*, j_*)$ can also be represented by the matrix

$$ \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} : \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} $$

Now, you get a short exact sequence from that piece of the MVS in the standard way:

$$ 0\longrightarrow \mathrm{ker}\ \partial \longrightarrow H_1(\mathrm{T}^2) \longrightarrow \mathrm{im}\ \partial \longrightarrow 0 $$

But, on one hand,

$$ \mathrm{im}\ \partial = \mathrm{ker}\ (i_*, j_*) = \mathrm{\ker}\ \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \mathbb{Z} $$

On the other hand,

$$ \mathrm{ker}\ \partial = \mathrm{im}\ (k_*-l_*) =(\mathbb{Z}\oplus \mathbb{Z}) / \mathrm{ker}(k_*-l_* ) = (\mathbb{Z}\oplus \mathbb{Z})/\mathrm{im}(i_*, j_*) = \mathbb{Z} $$

Hence, we have the following short exact sequence:

$$ 0 \longrightarrow \mathbb{Z} \longrightarrow H_1(\mathbb{T}^2) \longrightarrow \mathbb{Z} \longrightarrow 0 \ . $$


$$ H_1(\mathbb{T}^2) = \mathbb{Z} \oplus \mathbb{Z} \ . $$

EDIT. Some more details. Inclusions

$$ A \stackrel{i}{\longleftarrow} A\cap B \stackrel{j}{\longrightarrow} B $$

induce morphims

$$ H_1(A) \stackrel{i_*}{\longleftarrow} H_1(A\cap B ) \stackrel{j_*}{\longrightarrow} H_1(B) \ . $$

Hence a morphism

$$ (i_*, j_*) : H_1(A\cap B) \longrightarrow H_1(A) \times H_1(B) = H_1(A) \oplus H_1(B) \ , $$

whose value on $1$-cycles $\sigma : \Delta^1 \longrightarrow A\cap B$ is

$$ (i_*, j_*) [\sigma] = (i_*[\sigma], j_* [\sigma]) = ([i\circ \sigma],[j\circ \sigma]) . $$

Hence, if $a: \Delta^1 \longrightarrow A\cap B$ and $b: \Delta^1 \longrightarrow A\cap B$ are $1$-cycles representatives of the homology classes $\alpha$ and $\beta$, respectively, you have

$$ (i_*,j_*) (\alpha, 0) = (i_*,j_*) [a] = ([i\circ a],[j\circ a]) = (\alpha, \alpha) = (\alpha , \beta) $$

and analogously for $\beta$.

EDIT2. More details. To break a long exact sequence

$$ \dots \stackrel{f_{i-2}}{\longrightarrow} A_{i-1} \stackrel{f_{i-1}}{\longrightarrow} A_i \stackrel{f_i}{\longrightarrow} A_{i+1} \stackrel{f_{i+1}}{\longrightarrow} \dots $$

in "the standard way" means that, at any point, you can get short exact sequences like

$$ 0 \longrightarrow \mathrm{ker}\ f_i \longrightarrow A_i \longrightarrow \mathrm{im}\ f_i \longrightarrow 0 $$

where the arrow on the left is the inclusion and that on the right just $f_i$.

  • $\begingroup$ OK, I think I start to understand: you can write the map as this matrix because you write an element $\alpha$ in $H_n(A\cap B)$ as $(\alpha, 0)$ even though the elements in $H_1(A\cap B)$ are not actually pairs. Well they kind of are because $H_1(A \cap B) \cong \mathbb{Z} \oplus \mathbb{Z}$ $\endgroup$ Aug 19, 2011 at 10:33
  • $\begingroup$ Can I ask you one more question: Dylan mentioned orientation. Where does that come in? I think neither of you uses it anywhere.... $\endgroup$ Aug 19, 2011 at 10:39
  • $\begingroup$ Well, they're, since $H_1(A\cap B) = \mathbb{Z}\oplus \mathbb{Z}$. As for orientations, the only difference would be if you want $\alpha = \beta$ or $\alpha = -\beta$. $\endgroup$ Aug 19, 2011 at 13:22
  • 1
    $\begingroup$ What I did. :-) -See my second addition to my answer. $\endgroup$ Aug 19, 2011 at 16:58
  • 1
    $\begingroup$ @AgustíRoig This is a superb answer I wish I could give it more than 1 upvote. $\endgroup$
    – user38268
    Oct 21, 2012 at 13:17

$\newcommand{\Z}{\mathbf{Z}}$You can do this, but you need to know something about the maps — the exactness of the sequence is wonderful, but that often won't compute everything on its own. Anyway, if I want to get $H_1(T)$ then I write out $$ H_1(A \cap B) \stackrel{f}{\to} H_1(A) \oplus H_1(B) \to H_1(T) \to H_0(A \cap B) \stackrel{g}{\to} H_0(A) \oplus H_0(B). $$ We're going to replace the outer terms using certain isomorphisms, but let's think carefully about those. The picture is

A Mayer-Vietoris decomposition for the torus

If we view $A$ as $S^1 \times [0, 1]$, then the projection $p\colon S^1 \times [0, 1] \to S^1 \times \{1\}$ onto the top circle is a homotopy equivalence. Thus it induces an isomorphism $p_*$ on homology. Picking a generating loop for $H_1(S^1 \times \{1\})$, as shown, yields an isomorphism $H_1(A) \to \Z$. Any loop in $A$ whose projection is that generator, with the same orientation, will map to $1 \in \Z$.

I'll let you finish this program, using the orientation on $B$ shown and the like orientations on the two components of $A \cap B$. The upshot is that if we make these identifications and view our sequence as $$ \Z \oplus \Z \stackrel{f}{\to} \Z \oplus \Z \to H_1(T) \to \Z \oplus \Z \stackrel{g}{\to} \Z \oplus \Z, $$ then $f$ is given by $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$; after changing bases, this is $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$. It's easier to check that $g$ can be expressed by the same matrix. We obtain a short exact sequence $$ 0 \to \operatorname{Coker} f \to H_1(T) \to \operatorname{Ker} g \to 0. $$ We know that the outside terms are both isomorphic to $\mathbf{Z}$; since $\Z$ is free, it follows that $H_1(T) \approx \Z \oplus \Z$.

Getting $H_2 \approx \Z$ is similar, but of course $H_2(A)$, $H_2(B)$, and $H_2(A \cap B)$ are all trivial. Try the Klein bottle when you're done!

  • $\begingroup$ This is a little sketchy (I haven't thought about this stuff in a while), but I thought you should get some answer. Let me know if it's confusing. $\endgroup$ Aug 19, 2011 at 0:04
  • $\begingroup$ Thank you! Where do you use the orientation? $\endgroup$ Aug 19, 2011 at 12:46
  • $\begingroup$ And where does $0 \rightarrow Coker f \rightarrow H_1(T) \rightarrow Ker g \rightarrow 0$ come from? $\endgroup$ Aug 19, 2011 at 12:47
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    $\begingroup$ @Matt This is the trick of splitting long exact sequences up into short ones. Here $H_1(T)$ surjects onto the kernel of $g$, and the kernel of that map is the image of $H_1(A) \oplus H_1(B)$ in $H_1(T)$, which is isomorphic to $H_1(A) \oplus H_1(B)$ mod the kernel of $H_1(A) \oplus H_1(B) \to H_1(T)$, and this kernel is the image of $f$. $\endgroup$ Aug 19, 2011 at 15:46
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    $\begingroup$ @Matt For the orientation I'm a little less sure. I wanted to be careful in order to not compute the homology of the Klein bottle (that would be a good next exercise in Mayer-Vietoris), for one thing. It seems important for the top and bottom of $A$ to be consistently oriented (similarly for $B$), but if I flip around the orientations on $B$ I don't think anything bad happens. $\endgroup$ Aug 19, 2011 at 16:17

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