Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 21F (Algebraic Topology)

Suppose $X=A\cup B$ and that the interiors of $A$ and $B$ cover $X$. The Mayer–Vietoris theorem gives the long exact sequence in homologywhere all four displayed maps except the connecting homomorphism $\partial$ are induced by the relevant inclusion maps.

To prove it, let $C_{\ast }^{A,B}(X)$ be the chain complex generated by those singular simplices whose images lie wholly in $A$ or wholly in $B$. The sequenceis a short exact sequence of chain complexes. Repeated barycentric subdivision makes every singular simplex small enough to lie in $A$ or $B$, and the subdivision operator is chain homotopic to the identity. The inclusion $C_{\ast }^{A,B}(X)\hookrightarrow C_{\ast }(X)$ therefore induces an isomorphism on every homology group. The long exact homology sequence of a short exact sequence of chain complexes now gives the displayed sequence. Explicitly, if an $n$-cycle $z=a+b$ with $a\in C_n(A)$ and $b\in C_n(B)$, then $\partial a=-\partial b$ lies in $C_{n-1}(A\cap B)$ and represents the connecting class $\partial [z]$.

Write the closed oriented surface of genus $g$ as ${\Sigma }_g=A\cup B$, where $B$ is a closed disk, $A$ is a slightly enlarged once-punctured surface, and $A\cap B$ is an annulus. Thus $B$ is contractible, $A\cap B$ is homotopy equivalent to the circle, and $A$ is homotopy equivalent to the wedge sum of $2g$ circles. The boundary circle represents the product of the $g$ commutators in the fundamental group of $A$, so its image in the first homology group, which is the abelianization of that group, is zero.

The relevant part of the Mayer–Vietoris sequence is consequentlyThe degree-zero part says that the connected space ${\Sigma }_g$ has $H_0({\Sigma }_g)\cong \mathbb{Z}$, while the groups above degree two vanish. Hence

Let $T$ be the surface with boundary obtained by deleting the interiors of the $n$ disks. By excision,so $H_2(S,T)\cong {\mathbb{Z}}^n$ and the other relative groups vanish. In the long exact sequence in relative homology, the mapsends an orientation class to $(1,\dots ,1)$, up to an overall choice of sign. It is therefore injective and has cokernel ${\mathbb{Z}}^{n-1}$. Exactness gives $H_2(T)=0$ and a short exact sequenceThis sequence splits because its right-hand group is a free abelian group. Since $T$ is connected, we obtain

No. Here $T$ is the pair of pants, and part (b) gives $H_2(T)=0$, whereasIf $g:T^2\to T$ made $f\circ g$ homotopic to the identity map of the torus, functoriality of homology and homotopy invariance of homology would giveon $H_2(T^2)$. But $g_{\ast }$ factors through the zero group $H_2(T)$, so the left side is zero while the right side is the identity on $\mathbb{Z}$, a contradiction.