Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2009 / B / 6

A closed manifold is a compact manifold without boundary. Here $M$ is Hausdorff and every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$; the usual countability condition is included in the manifold convention. Compactness also gives a finite atlas and hence second countability.

A local orientation at $x$ is a choice of generator of $H_n(M,M\setminus \{x\};\mathbb{Z})\cong \mathbb{Z}$. An orientation of a manifold is a compatible choice of these generators as $x$ varies: locally they are induced by the same coordinate-ball orientation, or equivalently by a local relative homology class. For an oriented closed manifold, the fundamental class $[M]\in H_n(M;\mathbb{Z})$ is the unique class mapping to the chosen generator at every point. On disconnected closed manifolds this is the sum of the fundamental classes of their finitely many components.

For a triangulated or Delta complex description, orient the top-dimensional simplices compatibly with the local orientation and sum them with their corresponding signs. Each codimension-one face appears twice with opposite induced orientations, so the sum is a cycle. At a point inside any top simplex its image is the prescribed local generator, so this cycle represents $[M]$. In the connected case $H_n(M)=\mathbb{Z}$ generated by this class.

The homology conclusions below use the opening closed-manifold hypothesis for $M_1,M_2$. That hypothesis is necessary if their later description is read independently: removing a ball from ${\mathbb{R}}^n$ for $n\ge 2$ creates a nonzero $(n-1)$st homology group, so the claimed isomorphism would fail without closedness.

Remove the ball interiors. Each remaining piece inherits an orientation of a manifold and hence a boundary orientation. Choose the orientation on the second piece, reversing its original choice if necessary, so that the gluing map identifies these boundary orientations oppositely. Its two possible orientation signs are accommodated by this choice.

Near a glued point, the two half-space collar charts combine to give a full Euclidean chart. The opposite induced boundary orientations make the two interior orientations agree in this chart. Away from the seam, keep the orientations already chosen on the two pieces. These fit together globally, so the connected sum is orientable. This does not require that an arbitrarily chosen gluing preserve both initially specified orientations unchanged.

Write $M$ for either connected closed oriented manifold and $D$ for its embedded closed ball. A precise relative form of Poincare duality for a compact locally contractible subset isThis is Poincaré-Alexander duality. It applies to the printed closed-ball complement without replacing $D$ by its interior. Since $D$ is contractible, the relative group is $\mathbb{Z}$ at $j=n$ and zero otherwise. Naturality identifies $H_n(M)\to H_n(M,M\setminus D)$ with restriction $H^0(M)\to H^0(D)$, which is an isomorphism because both spaces are connected. Equivalently, the fundamental class restricts to the relative orientation generator on the ball.

In the long exact sequence of a pair, the only potentially exceptional part is thereforeExactness gives $H_n(M\setminus D)=0$ and the desired isomorphism in degree $n-1$. In every lower degree both adjacent relative terms are zero, so the same exact sequence gives the remaining isomorphisms. ThusFor the pieces $M_i\setminus \mathrm{int}{D}_i$ used in the gluing, the same conclusion follows from excision and the pair $(D^n,S^{n-1})$, or from the radial deformation of a punctured ball to its boundary. These pieces also have zero top homology, and the image of their boundary sphere's fundamental class is zero: it is the boundary of their relative oriented fundamental chain.

Let $N=M_1\#M_2$ and $A_i=M_i\setminus \mathrm{int}{D}_i$. Use open thickenings of the two pieces, whose intersection deformation retracts onto $S^{n-1}$. The Mayer-Vietoris sequence then has terms from $H_k(A_1)\oplus H_k(A_2)$ and $H_k(S^{n-1})$.

For $n\ge 2$, the pieces and intersection are connected. The map on $H_0$ from the intersection is $1\mapsto (1,-1)$ and is injective; it gives $H_0(N)=\mathbb{Z}$ and contributes no extra degree-one class. In intermediate degrees the sphere has no homology, except potentially at $n-1$. At that degree its map to $H_{n-1}(A_1)\oplus H_{n-1}(A_2)$ is zero because each oriented boundary sphere bounds the punctured piece's relative fundamental chain. Using $H_n(A_i)=0$, the top part of the sequence therefore gives $H_n(N)\cong \mathbb{Z}$ and $H_{n-1}(N)\cong H_{n-1}(A_1)\oplus H_{n-1}(A_2)$. The $H_0$ injection just noted handles the adjacent term when $n=2$.

Together with part (ii), the answer isFor $n=1$, each connected closed manifold is a circle and each punctured piece is an interval. The intersection is $S^0$ with two components: its $H_0$ map to the two interval components has rank one and kernel $\mathbb{Z}$. Mayer-Vietoris gives $H_1(N)=H_0(N)=\mathbb{Z}$, so the same displayed formula holds, with no intermediate degrees. The usual connected-sum construction here uses positive dimension.