# 19. Construction of the Mayer-Vietoris sequence

Video (All of Section 19)

This section will constitute the proof of Theorem 15.1. As before, let $X$ be a topological space, and let $U$ and $V$ be open sets with $X=U\cup V$. We name the inclusion maps as shown below:

This gives a sequence of chain maps as follows:

 $C_{*}(U\cap V)\xrightarrow{\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}C_{*}(U)\oplus C_{*}(V)\xrightarrow{\left[% \begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]}C_{*}(U\cup V).$

If this was a short exact sequence, then we could apply Theorem 17.2 to get the Mayer-Vietoris sequence. Unfortunately, however, this is not quite a short exact sequence: we will see that the first map is injective and the image of the first map is the kernel of the second one, but the second map is not surjective. We will need an extra step involving subdivision to deal with this issue.

###### Definition 19.1.

Put

 $\displaystyle S^{\prime}_{n}(U)$ $\displaystyle=\{u\colon\Delta_{n}\to U\;|\;u(\Delta_{n})\not\subseteq U\cap V\}$ $\displaystyle S^{\prime}_{n}(V)$ $\displaystyle=\{u\colon\Delta_{n}\to V\;|\;u(\Delta_{n})\not\subseteq U\cap V\}$ $\displaystyle S^{\prime}_{n}(X)$ $\displaystyle=\{u\colon\Delta_{n}\to X\;|\;u(\Delta_{n})\not\subseteq U\text{ % and }u(\Delta_{n})\not\subseteq V\}$ $\displaystyle S_{n}(U,V)$ $\displaystyle=\{u\colon\Delta_{n}\to X\;|\;u(\Delta_{n})\subseteq U\text{ or }% u(\Delta_{n})\subseteq V\}.$

We also define $C^{\prime}_{n}(U)={\mathbb{Z}}\{S^{\prime}_{n}(U)\}$ and similarly for $C^{\prime}_{n}(V)$, $C^{\prime}_{n}(X)$ and $C_{n}(U,V)$.

###### Remark 19.2.

If $u\in S^{\prime}_{n}(U)$ then $u$ does not send the whole of $\Delta_{n}$ into $U\cap V$, but it may send some faces of $\Delta_{n}$ into $U\cap V$. Because of this, $C^{\prime}_{*}(U)$ need not be closed under $\partial$, so it need not be a subcomplex of $C_{*}(U)$. However, this will not matter for our immediate purposes.

We now note that

 $\displaystyle S_{n}(U)$ $\displaystyle=S_{n}(U\cap V)\cup S^{\prime}_{n}(U)$ $\displaystyle S_{n}(V)$ $\displaystyle=S_{n}(U\cap V)\cup S^{\prime}_{n}(V)$ $\displaystyle S_{n}(U,V)$ $\displaystyle=S_{n}(U\cap V)\cup S^{\prime}_{n}(U)\cup S^{\prime}_{n}(V)$ $\displaystyle S_{n}(U\cup V)$ $\displaystyle=S_{n}(U\cap V)\cup S^{\prime}_{n}(U)\cup S^{\prime}_{n}(V)\cup S% ^{\prime}_{n}(X),$

and all these unions involve disjoint sets. It follows that

 $\displaystyle C_{n}(U)$ $\displaystyle=C_{n}(U\cap V)\oplus C^{\prime}_{n}(U)$ $\displaystyle C_{n}(V)$ $\displaystyle=C_{n}(U\cap V)\oplus C^{\prime}_{n}(V)$ $\displaystyle C_{n}(U,V)$ $\displaystyle=C_{n}(U\cap V)\oplus C^{\prime}_{n}(U)\oplus C^{\prime}_{n}(V)$ $\displaystyle C_{n}(U\cup V)$ $\displaystyle=C_{n}(U\cap V)\oplus C^{\prime}_{n}(U)\oplus C^{\prime}_{n}(V)% \oplus C^{\prime}_{n}(X),$

Thus, in our earlier sequence, the map $\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]\colon C_{n}(U\cap V)\to C_{n}(U)\oplus C_{n}(V)$ becomes the map

 $C_{n}(U\cap V)\to C_{n}(U\cap V)\oplus C^{\prime}_{n}(U)\oplus C_{n}(U\cap V)% \oplus C^{\prime}_{n}(V)$

given by $a\mapsto(a,0,-a,0)$. Similarly, the map $\left[\begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]\colon C_{n}(U)% \oplus C_{n}(V)\to C_{n}(U\cup V)$ becomes the map

 $C_{n}(U\cap V)\oplus C^{\prime}_{n}(U)\oplus C_{n}(U\cap V)\oplus C^{\prime}_{% n}(V)\to C_{n}(U\cap V)\oplus C^{\prime}_{n}(U)\oplus C^{\prime}_{n}(V)\oplus C% ^{\prime}_{n}(X)$

given by $(a,b,a^{\prime},c)\mapsto(a+a^{\prime},b,c,0)$. From this it is clear that we have a short exact sequence

 $C_{*}(U\cap V)\xrightarrow{\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}C_{*}(U)\oplus C_{*}(V)\xrightarrow{\left[% \begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]}C_{*}(U,V),$

and an inclusion $C_{*}(U,V)\to C_{*}(X)$ of chain complexes. By applying Theorem 17.2 to the short exact sequence, we get something which is essentially the Mayer-Vietoris sequence except that it involves $H_{*}(C_{*}(U,V))$ instead of $H_{*}(U\cup V)=H_{*}(X)$. To complete the construction of the Mayer-Vietoris sequence, we need to show that $H_{*}(C_{*}(U,V))$ is actually the same as $H_{*}(X)$.

###### Lemma 19.3.

For any $u\in C_{n}(X)$ there exists $k\geq 0$ such that $\operatorname{sd}^{k}(u)\in C_{n}(U,V)$.

###### Proof.

We can easily reduce to the case where $u$ is a single map $\Delta_{n}\to X$. Put $U^{\prime}=u^{-1}(U)$ and $V^{\prime}=u^{-1}(V)$, so $U^{\prime}$ and $V^{\prime}$ are open in $\Delta_{n}$ with $U^{\prime}\cup V^{\prime}=\Delta_{n}$. This means that $\{U^{\prime},V^{\prime}\}$ is an open cover of $\Delta_{n}$, so Proposition 8.31 tells us that there is a Lebesgue number, say $\epsilon>0$. The identity chain $\iota_{n}\in C_{n}(\Delta_{n})$ has diameter $\sqrt{2}$, so $\operatorname{sd}^{k}(\iota_{n})$ has diameter at most $(n/(n+1))^{k}\sqrt{2}$. If we choose $k$ large enough, then this diameter will be less than $\epsilon$, and it will follow that every simplex involved in $\operatorname{sd}^{k}(\iota_{n})$ is either contained in $U^{\prime}$ or contained in $V^{\prime}$. It follows that every simplex involved in the chain $\operatorname{sd}^{k}(u)=u_{*}(\operatorname{sd}^{k}(\iota_{n}))$ is either contained in $U$ or contained in $V$, so $\operatorname{sd}^{k}(u)\in C_{n}(U,V)$ as claimed. ∎

###### Corollary 19.4.

The homology of the quotient complex $Q_{*}=C_{*}(X)/C_{*}(U,V)$ is zero.

###### Proof.

First, note that the subdivision map $\operatorname{sd}\colon C_{*}(X)\to C_{*}(X)$ sends $C_{*}(U)$ to $C_{*}(U)$ and $C_{*}(V)$ to $C_{*}(V)$ so it also sends the subcomplex $C_{*}(U,V)=C_{*}(U)+C_{*}(V)$ to itself. We therefore have an induced map $\operatorname{sd}\colon Q_{*}\to Q_{*}$ given by $\operatorname{sd}(u+C_{n}(U,V))=\operatorname{sd}(u)+C_{n}(U,V)$. Similarly, the chain homotopy $\sigma\colon C_{n}(X)\to C_{n+1}(X)$ induces a chain homotopy $\sigma_{n}\colon Q_{n}\to Q_{n+1}$. We showed previously that $\partial\sigma+\sigma\partial=\operatorname{id}-\operatorname{sd}$ on $C_{*}(X)$, and it follows that we have the same relation on $Q_{*}$. We therefore deduce from Proposition 14.7 that the map $\operatorname{sd}_{*}\colon H_{n}(Q_{*})\to H_{n}(Q_{*})$ is the identity. Consider an element $q\in H_{n}(Q_{*})$. This has the form $q=z+B_{n}(Q_{*})$ for some $z\in Z_{n}(Q_{*})\leq Q_{n}=C_{n}(X)/C_{n}(U,V)$. This in turn has the form $z=u+C_{n}(U,V)$ for some $u\in C_{n}(X)$. For sufficiently large $k$ we have $\operatorname{sd}^{k}(u)\in C_{n}(U,V)$, so $\operatorname{sd}^{k}(z)=0$, so $\operatorname{sd}_{*}^{k}(q)=0$. As $\operatorname{sd}_{*}$ is the identity this means that $q=0$. Thus, we have $H_{n}(Q_{*})=0$ as claimed. ∎

###### Corollary 19.5.

The inclusion $C_{*}(U,V)\to C_{*}(X)$ induces an isomorphism $H_{*}(C_{*}(U,V))\to H_{*}(X)$.

###### Proof.

We have a short exact sequence of chain complexes $C_{*}(U,V)\to C_{*}(X)\to Q_{*}$. Theorem 17.2 therefore gives us exact sequences

 $H_{n+1}(Q_{*})\to H_{n}(C_{*}(U,V))\to H_{n}(C_{*}(X))\to H_{n}(Q_{*}).$

The first and last groups are zero by Corollary 19.5, so exactness forces the middle map to be an isomorphism. ∎

This completes the construction of the Mayer-Vietoris sequence.