Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2009 / B / 4

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.