I'm reading Hatcher's book "Algebraic topology", the section about Mayer-Vietoris sequences (p150). It is claimed that there is also a version for reduced homology. However, let $X$ be space and $\epsilon: C_0(X) \to \mathbb{Z}$ the canonical map. It appears to me that we still need to prove that $$\frac{\operatorname{Ker}(\epsilon: C_0(A+B) \to \mathbb{Z})}{\operatorname{Im}(\partial_1: C_1(A + B) \to C_0(A+B))}\cong \widetilde{H}_0(X)= \frac{\operatorname{Ker}(\epsilon: C_0(X) \to \mathbb{Z})}{\operatorname{Im}(\partial_1: C_1(X) \to C_0(X))}$$ to obtain the version of the Mayer-Vietoris sequence for reduced homology. Would it be correct to say that this follows by modifying the proof of the excision theorem (prop 2.21)?

Thanks in advance for the help!


2 Answers 2


Given a short exact sequence of chain complexes $$0 \to \mathbf A \stackrel{\mathbf i}{\longrightarrow} \mathbf B \stackrel{\mathbf j}{\longrightarrow} \mathbf C \to 0$$ we get a long exact sequence of homology groups $$\ldots \to H_n(\mathbf A) \stackrel{\mathbf i_*}{\longrightarrow} H_n(\mathbf B) \stackrel{\mathbf j_*}{\longrightarrow} H_n(\mathbf C) \stackrel{\partial}{\longrightarrow} H_{n-1}(\mathbf A) \stackrel{\mathbf i_*}{\longrightarrow} H_{n-1}(\mathbf B) \to \ldots$$ See Hatcher Theorem 2.16.

Now recall your question Chain homotopy definition . In the answer we have seen that there is a more general concept of chain complex allowing also non-zero groups in negative dimensions. Theorem 2.16 remains true for short exact sequences of such more general chain complexes; you do not even need to adapt the proof.

An example is the augmented singluar chain complex which has an additional $C_{-1}(X) = \mathbb Z$ (and $C_n(X) = 0$ for $n < -1$). Using this chain complex we get the long exact sequence for reduced homology groups.

The usual Mayer-Vietoris sequence is obtained by applying this to the short exact sequence $$ 0 \to C_*(A \cap B) \stackrel{\varphi}{\longrightarrow} C_*(A) \oplus C_*(B) \stackrel{\psi}{\longrightarrow} C_*^{\mathcal U}(X) \to 0$$ with $\mathcal U = \{A,B\}$ and using the fact that $\iota : C_*^{\mathcal U}(X) \hookrightarrow C_*(X)$ is a chain homotopy equivalence (Proposition 2.21).

To get the Mayer-Vietoris sequence for reduced homology groups we shall show that Proposition 2.21 is also true for the augmented chain complexes (which we denote by $C_\#$); this means that the "augmented map" $\iota : C_\#^{\mathcal U}(X) \hookrightarrow C_\#(X)$ is a chain homotopy equivalence. What does this map look like in dimension $-1$?

Note that $C_0^{\mathcal U}(X) = C_0(X)$ because each singular $0$-simplex automatically maps into some $U_i \in \mathcal U$. Thus $\iota = id$ in dimension $0$ and we can augment $\iota$ with $id: \mathbb Z \to \mathbb Z$ in dimension $-1$.

On p. 123 Hatcher defines a chain map $\rho : C_*(X) \to C_*^{\mathcal U}(X)$ and a chain homotopy $D$ from $\mathbb I$ to $\iota \rho$ and moreover shows that $\rho \iota = \mathbb I$.

We conclude that $\rho = id$ in dimension $0$ because $\iota = id$ in dimension $0$. Therefore we can augment $\rho$ with $id: \mathbb Z \to \mathbb Z$ in dimension $-1$. These augented maps trivially satisfy $\rho \iota = \mathbb I$ on $C_\#^{\mathcal U}(X)$.

We next show that Hatcher's chain homotopy $D$ augments to a chain homotopy from $\mathbb I$ to $\iota \rho$ on $C_\#(X)$; this will prove our modification of Proposition 2.21. Recall that $D_{-1} : 0 \to C_0(X)$ for Hatcher's original $D$; thus $$\partial D_0 = \partial D_0 + D_{-1}\partial = id_0 - \iota_0 \rho_0 .$$ But $\iota_0 \rho_0 = id_0$, thus $\partial D_0 = 0$. Therefore we can augment $D$ as follows to $C_\#(X)$:

  • Take $\bar D_{-1} : \mathbb Z \to C_0(X)$ and $\bar D_{-2} : 0 \to \mathbb Z$ the zero maps.

Then $\partial D_0 + \bar D_{-1}\varepsilon = 0 = id_0 - \iota_0 \rho_0$ and $\varepsilon \bar D_{-1} + \bar D_{-2}0 = 0 = id_{\mathbb Z} - \iota_{-1} \rho_{-1}$.

  • $\begingroup$ Thanks for your answer. I see this, but I think you misunderstood the question! Rather, if you consider the augmented chain complex $\dots \to C_1(A+B) \to C_0(A+B) \to \mathbb{Z}\to X$ and look at the long exact sequence associated to this complex, you encounter the associated homology group $\operatorname{Ker}(C_0(A+B)\to \mathbb{Z})/\operatorname{Im}(\partial_1: C_1(A+B) \to C_0(A+B))$. We still need to justify that this coincides with $\widetilde{H}_0(X)$ and this is exactly what my question is about. Please let me know if it is clear now or if I should further clarify. $\endgroup$
    – user839372
    Jun 17, 2021 at 15:35
  • $\begingroup$ There is a typo in the above comment: $\mathbb{Z}\to X$ should be $\mathbb{Z}\to 0$. $\endgroup$
    – user839372
    Jun 17, 2021 at 18:15
  • $\begingroup$ @user839372 See my update. $\endgroup$
    – Paul Frost
    Jun 17, 2021 at 22:38

I like the approach to Mayer-Vietoris that Hatcher presents in the exercises. Exercise 38 on p. 159 says that a commutative diagram $$ \begin{array}{ccccccc} \dots & \to & C_{n+1} & \to & A_n & \to & B_n & \to & C_n & \to & \dots \\ & & \downarrow & & \parallel & & \downarrow & & \downarrow \\ \dots & \to & E_{n+1} & \to & A_n & \to & D_n & \to & E_n & \to & \dots \end{array} $$ (where the rows are exact and every third vertical map is an isomorphism) yields a long exact sequence $$ \dots \to E_{n+1} \to B_n \to C_n \oplus D_n \to E_n \to \dots $$ Apply this to long exact sequences for the pairs $(A, A \cap B)$ and $(X, B)$: inclusion maps give maps between the sequences with excision providing the isomorphisms in every third spot. This works with either reduced or unreduced homology since you have long exact sequences in both situations and excision doesn't care about reduced vs. unreduced.

  • $\begingroup$ Thanks for the alternative view! But do you think my approach can work too? $\endgroup$
    – user839372
    Jun 17, 2021 at 17:03
  • $\begingroup$ One of the points behind my post, and I think @PaulFrost's post, too, is that Mayer-Vietoris should be a formal consequence of the long exact sequence for a pair and excision. So you shouldn't have to do such precise analysis. $\endgroup$ Jun 17, 2021 at 17:55
  • $\begingroup$ Yes, but the excision part is unclear in Hatcher's approach, hence my question. $\endgroup$
    – user839372
    Jun 17, 2021 at 18:14
  • $\begingroup$ I claim that Hatcher's Proposition 2.21 holds for augmented chain complexes, and it's almost immediate: just extend the chain homotopy on the ordinary chain complex to the augmented one by using the zero map, as Hatcher does for $LC$ on p. 122. (If two chain maps are chain homotopic and if they both are extended by the same map on the augmented part, then you can extend the chain homotopy this way.) $\endgroup$ Jun 17, 2021 at 18:37
  • $\begingroup$ Yes, something along this line is what I meant when I wrote "Modify the proof of 2.21". Thanks a lot for the help! $\endgroup$
    – user839372
    Jun 17, 2021 at 20:34

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