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A map with two preimages but degree zero

Statement refuted

The unsigned number of points in a regular fibre need not equal the degree. The smooth proper map F:S1→S1 given by F(eiθ)=eisin⁡θ has the regular value 1 with exactly two preimages and opposite local signs, hence degree zero.

Facts & Assumptions

Given: Both circles have their counterclockwise orientations.

[F1]

[t]↦(cos⁡2πt,sin⁡2πt) is a homeomorphism from R/Z to the unit circle identifies the quotient coordinate [t]∈R/Z with e2πit.

[F5]

Regular-value formula for compact-support degree computes degree as the signed sum over any supplied regular fibre.

Counterexample

1.1F1F2F3F4given

Under [F1] the displayed map is G([t])=[sin⁡(2πt)2π]. It is well defined because replacing t by an integer translate does not change the sine by [F2]. In local increasing angular coordinates its lifts differ only by integer constants and have derivative cos⁡(2πt) by [F3]; repeated differentiation cycles through sine and cosine, so G is smooth. If K is compact in the target, [F4] makes K closed, hence G−1(K) closed in the compact source and therefore compact. Thus G, equivalently F, is proper.

2.1F2F3F5step 1.1algebra

The fibre of [0], corresponding to 1∈S1, satisfies sin⁡(2πt)∈2πZ. Since ∣sin⁡(2πt)∣≤1<2π by [F2], this is equivalent to sin⁡(2πt)=0. The zero-set formula in [F2] gives exactly [t]=[0] or [t]=[1/2]. By [F3] their derivatives are respectively +1 and −1, so [0] is regular and [F5] gives deg⁡(F)=(+1)+(−1)=0, although this fibre has two points.

3.1F2F3F5step 1.1step 2.1∎

This witnesses the failed unsigned-count conclusion. For comparison, the target value [1/2] has empty fibre because every lifted value of G has absolute value at most 1/(2π)<1/2, and the empty regular-fibre sum again gives zero. The extreme target [1/(2π)] has the singleton preimage [1/4], but its derivative is cos⁡(π/2)=0, so it is critical rather than a counterexample to the regular-value formula. Quotient seams are handled by local lifts, and every fibre point used above is explicitly listed; no choice principle is used.

Remarks

The schematic marks the regular fibre over 1: the right and left source points have local signs +1 and −1, respectively (step 2.1). Both circles are oriented counterclockwise. The thick target arc is the image {eiu:−1≤u≤1}, not the whole target circle, since u=sin⁡θ. The arrow denotes F; both marked source points map to the marked target point. The proof above is independent of the illustration.

sourceS1targetS1µ=0+1µ=¼¡11F(eiµ)=eisinµthickarc:F(S1)#F¡1(1)=2;deg(F)=(+1)+(¡1)=0

Depends on

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