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

For $n\ge 1$, the integral singular homology of the sphere is $H_0(S^n)=\mathbb{Z}$, $H_n(S^n)=\mathbb{Z}$, and zero in other degrees. For $S^0$ there are two components, so $H_0(S^0)={\mathbb{Z}}^2$. Choose an orientation generator $[S^n]$ for positive dimension. The degree of a map is the integer satisfyingTo define a local degree of a continuous map at $x$ over $y=f(x)$, require $x$ to be isolated in that fibre. A small neighbourhood $U$ contains no other preimage of $y$, so $f$ maps $(U,U\setminus \{x\})$ to $(S^n,S^n\setminus \{y\})$. By excision their local top homology groups are both $\mathbb{Z}$, with generators fixed by the orientations. Multiplication by an integer in this induced map defines ${\mathrm{deg}}_{x}f$. Without fibre isolation, this particular local-degree definition is not in general available.

If $f^{-1}(y)=\{x_1,\dots ,x_r\}$ is finite, the local-degree formula isIndeed excision identifies $H_n(S^n,S^n\setminus f^{-1}(y))$ with the direct sum of the local groups at the $x_j$. The global fundamental class maps to the vector of their local orientation generators, namely $(1,\dots ,1)$. Naturality of the map of pairs sends that vector to ${\sum }_j{\mathrm{deg}}_{x_j}f$ in the local group at $y$. The other route through the naturality diagram sends the global class by $f_{\ast }$ first and therefore gives $\mathrm{deg}(f)$. The map from the target sphere's top homology to its local homology is an isomorphism, so the two integers agree. Empty fibres give degree zero by the same argument. Standard approximation by a homotopic smooth map permits the formula to be used with regular values when an original continuous map has no convenient finite fibre.

Identify $S^1$ with the unit complex numbers and put ${\alpha }_d(z)=z^d$ for any integer $d$, including the constant map for $d=0$. A lift of its angular coordinate to the real line is $t\mapsto dt$, so it sends the oriented circle's fundamental class to $d$ times itself. Thus $\mathrm{deg}({\alpha }_d)=d$.

The natural suspension isomorphism in reduced homology identifies ${\tilde{H}}_{j+1}(\Sigma X)$ with ${\tilde{H}}_j(X)$ and intertwines a map with its suspension. This isomorphism follows from Mayer-Vietoris sequence for the two contractible suspension cones, and its naturality gives the intertwining. Since $\Sigma S^j\cong S^{j+1}$, suspending a sphere self-map preserves degree. Repeating this construction givesThe positive-dimension convention is necessary for the arbitrary-degree assertion; self-maps of the two-point space do not realize arbitrary integers on reduced homology.

Let $a(x)=-x$ be the antipodal map, whose degree is $(-1{)}^{n+1}$. The invariance condition is $f\circ a=f$. Multiplicativity of degree gives $\mathrm{deg}(f)(-1{)}^{n+1}=\mathrm{deg}(f)$, so if $n$ is even,For odd $n$, factor the invariant map uniquely through the quotient $q:S^n\to {\mathbb{RP}}^n$, writing $f=gq$. The standard orientability and homology facts for Real projective space are that ${\mathbb{RP}}^n$ is orientable in odd dimension and has top integral homology $\mathbb{Z}$. With compatible orientation the double cover $q$ has degree two: both local sheets preserve orientation because the antipodal map does. Thus $g_{\ast }$ is multiplication by an integer $k$ and $\mathrm{deg}(f)=2k$.

Every such integer occurs. In the oriented projective space, choose a small oriented closed ball and collapse its exterior to a point. The resulting map $c:{\mathbb{RP}}^n\to S^n$ has degree one, because a point in the interior image has one orientation-preserving preimage. Composing it with a degree-$k$ sphere map from part (b) givesHere ${\alpha }_k$ denotes the dimension-$n$ map constructed by suspension. The factorization through $q$ makes the resulting map antipodally invariant.

A CW complex is built from a discrete set of zero-cells by attaching $k$-discs along maps of their boundary spheres into the previous skeleton. Its topology is the weak topology determined by the characteristic maps, and each cell closure meets only finitely many cells. The k-skeleton $X^k$ is the union of cells of dimension at most $k$, with $X^{-1}=\varnothing$.

Set $C_k^{CW}(X)=H_k(X^k,X^{k-1};\mathbb{Z})$. This is the free abelian group on the $k$-cells. The cellular boundary map is ${\partial }_k=q_{k-1}{\delta }_k$ with the maps in the question. In the long exact sequence of the pair $(X^{k-1},X^{k-2})$, the consecutive maps areExactness gives ${\delta }_{k-1}{q}_{k-1}=0$. HenceAt the lowest indices put negative chain groups equal to zero.

The cellular homology is $H_n^{CW}(X)=\ker {\partial }_n/\mathrm{im}{\partial }_{n+1}$. For a field $\mathbb{F}$, use $C_k^{CW}(X;\mathbb{F})=H_k(X^k,X^{k-1};\mathbb{F})$, the vector space with basis the $k$-cells, with the same connecting-and-quotient construction. Its homology is $H_n^{CW}(X;\mathbb{F})$. Equivalently tensor the integral free cellular chain complex with $\mathbb{F}$. The cellular homology theorem identifies these with singular homology using the same coefficients.

The identifications yield one vertex, $g$ one-cells and one two-cell, whose boundary word is $a_1^2\cdots a_g^2$. Thus the integral cellular chain complex, in degrees two, one and zero, hasFor $g\ge 1$ the first map is injective. The vector $(1,\dots ,1)$ is primitive and extends to an integral basis, so quotienting by twice that vector givesWith ${\mathbb{F}}_2=\mathbb{Z}/2\mathbb{Z}$, every boundary map vanishes. HenceFor cohomology, dualize the free cellular complex. The only nonzero possible coboundary over $\mathbb{Z}$ is ${\delta }^1:{\mathbb{Z}}^g\to \mathbb{Z}$, $(v_1,\dots ,v_g)\mapsto 2{\sum }_i{v}_i$. Its kernel is free of rank $g-1$ and its image is $2\mathbb{Z}$. ThusOver ${\mathbb{F}}_2$ the cochain maps are zero too, giving $H^0=H^2={\mathbb{F}}_2$, $H^1={\mathbb{F}}_2^g$, and zero higher groups. In particular the integral top homology and top cohomology differ, as their explicit complexes show. Finally the cell count gives

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.