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

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.