The count a fold cannot change

A linear map multiplies every area by the same number, the absolute value of its determinant, and the sign of the determinant says whether it turns the plane over. That is where the determinant began. A curved map does the same thing locally: near each point it is almost linear, the linear map that fits it best is its derivative, and the determinant of the derivative — the Jacobian determinant — says how much the map stretches area there and whether it flips it.

So far that is the determinant’s first property, applied one small square at a time. What is new is what happens to the whole. A linear map with non-zero determinant is one-to-one, and the area of an image is the area of the original times one number. A curved map can fold: it can carry one region of the plane on top of another, so that some points of the image are reached once and some are reached three times. The figure below is such a map, and every question in this essay is about the numbers printed beneath it.

The raw number of preimages jumps as a point moves: one here, three there. The claim is that if every preimage is counted with the sign of the determinant at it — plus one where the map keeps the plane’s orientation and minus one where it reverses it — the jumps cancel, and the count is the same at every point the figure marks. That signed count is the degree of the map, and the sign that makes it constant is the same sign the permanent threw away.

Which way round a small square comes back #

Before the count, the sign. Take a small square in the domain and number its corners anticlockwise. Push it through the map. If the map is nearly linear on the square — and on a small enough square it is — the image is nearly a parallelogram, with area close to the determinant times the square’s, and the corners come back in some order.

The top square sits where the determinant is positive, and its image is a slanted parallelogram with its corners still in anticlockwise order. The bottom square sits where the determinant is negative: its image has the corners in clockwise order, which is to say the map has turned that little piece of the plane over, the way a reflection does. The middle square straddles the curve where the determinant is nought, and its image is creased — the map folds it along that curve and lays one half on the other, so that almost no area is left.

That curve is the fold. On one side the map preserves orientation, on the other it reverses it, and along the curve itself it squashes a direction to nothing. A sheet of paper folded in half does exactly this, and the analogy is exact: the part folded over is face down, which is what a negative determinant is.

Hassler Whitney showed in 1955 that for a map of the plane to itself in general position, there are only two kinds of places where this happens. There are folds, like the curve above, and there are cusps, where a fold curve turns back on itself. Every more complicated singularity can be removed by an arbitrarily small change to the map, and these two cannot.

Three sheets or one, and the count that stays put #

The map in the hero is the standard cusp: $(u, v) \mapsto (u,\; v^3 + uv)$. It keeps the first coordinate and bends the second, and its Jacobian determinant is $3v^2 + u$, which is negative inside the sideways parabola $u = -3v^2$ and positive outside it. The image of that parabola is a sharp-pointed curve, the cusp, and inside the cusp every point has three preimages: the equation $v^3 + uv = y$ is a cubic in $v$, and for a point inside the cusp it has three real roots, for a point outside it one.

At the marked points in the figure the counts are three, three, one, one and one. Counted with signs they are $+1 - 1 + 1$, $+1 - 1 + 1$, $+1$, $+1$, $+1$ — one every time.

The reason is visible in the domain. As a target point moves toward the edge of the cusp, two of its three preimages move toward each other — one on the positive side of the fold, one on the negative side — and at the edge they meet on the fold and vanish together. They were a pair with opposite signs, so their disappearance changes the raw count by two and the signed count by nothing. Crossing the fold in the other direction creates a pair, again with opposite signs. A preimage can never appear or vanish alone, because in the domain the preimages move continuously and can only be created or destroyed where the determinant is nought, which is where the two sheets meet.

This is the whole mechanism, and it does not depend on the particular map. For any smooth map, the signed count of preimages can only change when the target point crosses the image of the domain’s boundary, because that is the only other place a preimage can come from or go to: entering or leaving the region across its edge.

The area formula, with the signs left in #

For a one-to-one map, the area of the image is the integral of the absolute Jacobian determinant over the domain. That is the change of variables every multiple integral uses, and it is the first property of the determinant written as a sum over small squares.

When the map folds, the integral of the absolute determinant still makes sense, but it no longer measures the image: a point reached three times contributes three times. Dropping the absolute value gives a different quantity, and a better one.

$$\int_{\text{domain}} \det Df \;=\; \int_{\text{target}} (\text{signed count of preimages}).$$

The argument is to cut the domain along the fold into pieces on which the map is one-to-one. Each piece maps onto part of the target, and the integral of the determinant over the piece is plus or minus the area of that part — plus if the piece is on the positive side of the fold, minus if it is on the negative side. Adding up the pieces adds up, for each target point, the signs of its preimages.

In the hero both sides are computed separately, by different routes, and printed in the caption. The determinant, integrated over the rectangle of the domain on a grid of 57,600 small squares, comes to about $4.447$. The signed count, computed at each point of a grid over the target by solving the cubic exactly and adding up the signs of the roots, integrates to about $4.434$. The two agree to the accuracy of the grids, and neither computation uses the other.

That is a genuinely surprising identity when stated for the whole target: the integral on the right counts the three-sheeted region inside the cusp once, exactly as if the fold had never happened.

A fold that counts for nothing #

The simplest fold is the one that takes the plane and lays the bottom half on the top half: $(x, y) \mapsto (x,\; y^2)$. Its determinant is $2y$, positive above the $x$-axis and negative below it.

Every point of the image has two preimages, one above the axis and one below, and their signs are opposite, so the signed count is nought everywhere. The integral of the determinant over the rectangle is nought as well — the positive half and the negative half cancel exactly. The image has area one and is covered twice, and the signed account of it is that it is not covered at all.

That is not a failure of the signed count; it is the point of it. A sheet of paper folded in half and pressed flat can be unfolded without tearing, and so can this map: it can be deformed, keeping the boundary away from any given target point, into a map that misses that point altogether. The signed count measures what cannot be undone by deformation, and a fold can always be undone. The raw count, which says two, is measuring something that can.

Squaring the disc, twice over #

Now a map with no folds at all. Squaring a complex number, $z \mapsto z^2$, written in coordinates is $(x, y) \mapsto (x^2 - y^2,\; 2xy)$, and its Jacobian determinant is $4(x^2 + y^2)$ — never negative, and nought only at the origin.

Every point of the image disc except the centre has two square roots, $w$ and $-w$, and both lie in the disc; both preimages count plus one; the signed count is two everywhere. The determinant integrated over the disc is $2\pi$, twice the disc’s area — the disc is laid down twice, both times face up.

This is the general pattern for a map that comes from a complex function. The derivative of such a map at a point is a rotation and a scaling, and the determinant of a rotation-and-scaling is the square of the scale factor, $|f'(z)|^2$, which cannot be negative. Maps from complex functions never fold. Their preimages all count plus one, so for them the signed count and the raw count are the same number.

The signed count is a winding number #

The signed count of a map on a disc is the same at every point it covers, and the only place it can change is across the image of the boundary circle. That suggests reading it off the boundary alone, and the reading is one made before.

As the boundary circle is traced once, its image winds round the centre some number of times — the winding number — and for every map in the figure that winding number is the signed count: $1$, $2$ and $3$ for the powers, and $-1$ for the reflection, whose determinant is negative everywhere. The two numbers are computed by routes that share nothing. One adds up the angle the boundary image turns through; the other integrates a determinant over the inside.

The identity has one consequence worth drawing out, because it is the reason the argument is worth making at all. Take a polynomial of degree $n$ with leading term $z^n$. On a large enough circle the polynomial is dominated by $z^n$, so its boundary image winds $n$ times round nought, and so its signed count at nought is $n$. It is a complex function, so it never folds, and every preimage counts plus one. So nought has preimages — $n$ of them, with multiplicity — and the polynomial has $n$ roots. That is the fundamental theorem of algebra, recovered from the fact that a determinant of the form $|f'|^2$ has no minus sign available.

A boundary that forces the middle to be covered #

The winding-number reading turns the signed count into a tool for proving that a map must reach a point, without finding the preimage.

Suppose a map of the disc leaves every point of the boundary circle where it was. Whatever it does inside — stretching, folding, crumpling — the image of the boundary is the circle itself, traced once anticlockwise, and it winds once round every point inside. So every inside point has signed count one, and a signed count of one cannot be made of no preimages at all. The map reaches every point of the disc. It cannot leave a hole.

That is the no-retraction theorem, and it is the heart of the result that every continuous map of a disc to itself leaves some point where it was. If a map had no fixed point, the ray from each image through its original point would carry the disc onto its own boundary circle while keeping the circle fixed — a map with a hole in its image at the centre, which the signed count has just forbidden. The fixed point is found by counting, not by searching.

The same counting, one dimension up, is why a sphere cannot be combed flat: a combing would let the identity map of the sphere be deformed into the antipodal map, and the two have signed counts of plus one and minus one. A whole number cannot change continuously, and the determinant’s sign is where the minus comes from — the antipodal map of an ordinary sphere reverses orientation, which is a negative determinant at every point.

What all three arguments share is that they never look inside. The boundary fixes the count; the count forces preimages; and the preimages themselves are never exhibited. That is the same economy the determinant always offered — a single number that knows about every configuration — and here the configurations are the sheets of a map rather than the terms of a sum.

Why the unsigned count could never do this #

It is worth being precise about what the sign bought, because it is the determinant’s alternating property doing its last job in a new setting.

The integral of $|\det Df|$ over the domain is the area of the image counted with multiplicity. It is a perfectly good number and it moves when the map is deformed: pull a fold deeper and the doubly covered region grows. The signed count at a point does not move under any deformation that keeps the boundary’s image away from that point, because it counts a region that has been folded over as covered once and then uncovered once. The first is the analogue of the permanent — every term positive, every configuration counted, nothing cancelling. The second is the analogue of the determinant, and the cancellation is what makes it an invariant: a quantity that deformation cannot change, and so one that can be computed from the simplest map in its class.

The difference is the same one the permanent turned on. There, cancellation made the determinant computable by elimination and its absence made the permanent intractable. Here, cancellation makes the signed count a whole number fixed by the boundary alone, and its absence leaves a quantity that depends on every detail of the map.

What the grids cannot settle #

The integrals in these figures are sums over grids of small squares and small cells, and they agree to two or three decimal places, not exactly. The exactness is in the preimage counts, which are found by solving a cubic or taking a square root in closed form and then checked by mapping each root forward. The theorem that the two integrals are equal is proved by the cutting argument above, not by the agreement of the grids.

The argument itself leans on two things the pictures do not show. One is smoothness: the map must be differentiable enough for the determinant to exist and to vary continuously, and a map built with corners can do things none of these can. The other is Sard’s theorem, which says that the points of the target reached from the fold — where a preimage has determinant nought and so no sign — make up a set of area nought. They are the images of the fold curves, thin curves in the pictures, and the signed count is simply not defined on them. The figure marks points off those curves, and a point exactly on the cusp’s edge would be a different story.

And the whole essay is in two dimensions. In three or more the Jacobian determinant is the volume factor of a box rather than a parallelogram, the fold becomes a surface, and the argument is the same, but no figure on a page can show a map of space folding over itself.

Still open: when a map with no folds is one-to-one #

A map whose Jacobian determinant is never nought has no folds, and so every preimage of every point counts with the same sign. Does it follow that each point has at most one preimage? For smooth maps, no: the exponential of a complex number has non-zero derivative everywhere and hits every non-zero point infinitely often. The question becomes sharp for polynomials.

In 1939 Ott-Heinrich Keller asked: if a map of the plane given by two polynomials has Jacobian determinant equal to a non-zero constant, must it be one-to-one, with an inverse that is also given by polynomials? This is the Jacobian conjecture. It has been checked for polynomials of low degree and proved under many extra assumptions, and several published proofs have turned out to be wrong. For real polynomial maps the stronger version — that a Jacobian determinant which merely never vanishes forces the map to be one-to-one — is false: Sergey Pinchuk gave a counterexample in 1994, a map of the plane with everywhere positive determinant that is not one-to-one.

Whether a constant determinant is enough, even for two polynomials in two variables, is not known. The determinant says the map never folds and never shrinks area; the question is whether that local information, in the rigid setting of polynomials, forces the global conclusion that no point is reached twice.