 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/26g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Assume first that

Y

is connected; otherwise the definition is made separately on each connected component. Choose a regular value

y \in Y

de g_{2} (f) = ∣ f^{- 1} (y) ∣ (mod 2) .

The following course results make this well-defined.

By Sard theorem, regular values are dense, so one can choose $y$ .
By the preimage theorem, $f^{- 1} (y)$ is a zero-dimensional submanifold because $dim X = dim Y$ . It is discrete and closed in compact $X$ , hence finite.
The parity is independent of the regular value. Indeed, after choosing a path between two regular values transverse to $f$ , its inverse image is a compact one-dimensional manifold whose boundary is the disjoint union of the two fibres. Every compact one-manifold has an even number of boundary points.

The same transverse-preimage argument applied to a smooth homotopy proves homotopy invariance. Consequently the definition depends only on

f

, rather than on the chosen regular value.

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