Brouwer’s fixed-point theorem

We prove that every continuous map from the closed unit ball to itself has a fixed point. The theorem is usually presented in topological language: a fixed-point-free map would produce a retraction of the ball onto its boundary, but the ball cannot retract onto its boundary. The proof below makes that mechanism visible. The key local fact is extremely simple. A smooth map into the sphere $S^{n-1}$ has derivative of rank at most $n-1$ , because every derivative vector must lie in the tangent hyperplane of the sphere. Consequently, such a map carries no infinitesimal $n$ -dimensional volume. A differential form measuring signed sphere-area then has closed pullback. Stokes’ theorem turns this local vanishing statement into a global boundary invariant.

For the circle, this invariant is total signed angle. In higher dimensions, it is total signed $(n-1)$ -dimensional sphere-area. The identity map of the boundary has positive total signed area, whereas a boundary map which extends through the ball has total signed area zero. A fixed-point-free self-map would force one boundary map to have both values at once.

We begin with the smooth case. The passage from smooth maps to continuous maps is handled at the end by moving the map slightly into the interior, extending it, and smoothing by convolution.

Write $\displaystyle B^n:={x\in\mathbb R^n:|x|\le1}, S^{n-1}:=\partial B^n.$ We shall prove that every continuous map $\displaystyle f:B^n\longrightarrow B^n$ has a fixed point.

Assume first that $f$ is smooth on a neighborhood of the closed ball, and suppose, towards a contradiction, that it has no fixed point. Then $\displaystyle x-f(x)\ne0\quad\text{for every }x\in B^n.$ Thus the displacement from $f(x)$ to $x$ has a well-defined direction. Forgetting its length and retaining only this direction gives a smooth map

$\displaystyle u:B^n\longrightarrow S^{n-1},\quad u(x):=\frac{x-f(x)}{|x-f(x)|}.$

This normalized displacement field is the essential object. It is defined on the whole ball precisely because $f$ is assumed fixed-point-free.

On the boundary sphere, $u$ is continuously deformable to the outward radial map $x\mapsto x$ . Indeed, for $x\in S^{n-1}$ and $0\le t\le1$ , consider

$\displaystyle H(x,t):=\frac{x-tf(x)}{|x-tf(x)|}.$

The denominator never vanishes. For if $x=tf(x)$ , then $\displaystyle 1=|x|=t|f(x)|\le t\le1.$ Every inequality would have to be equality. Thus $t=1$ and $|f(x)|=1$ , so $x=f(x)$ , contradicting the assumption. Hence $H$ is a smooth sphere-valued homotopy satisfying

$\displaystyle H(x,0)=x,\quad H(x,1)=u(x)\quad\text{for }x\in S^{n-1}.$

The contradiction will follow from two facts:

$\displaystyle \int_{S^{n-1}}u^*\alpha=0,$

because $u$ extends over the ball, while

$\displaystyle \int_{S^{n-1}}u^*\alpha=\int_{S^{n-1}}{\text{id}}^*\alpha={\text{Area}}(S^{n-1})>0,$

because $u|_{S^{n-1}}$ is homotopic through sphere-valued maps to the identity. The rest of the proof explains the form $\alpha$ and justifies these two identities directly from calculus.

2 dimensions

The case $n=2$ contains the entire idea in its most concrete form. The ball is the closed disk $D^2$ , its boundary is the circle $S^1$ , and the quantity being preserved is total signed angle. Write the standard coordinates on $\mathbb R^2$ as $(p,q)$ . On the circle define the one-form $\displaystyle \alpha:=p\,dq-q\,dp.$ Locally, a point of the circle can be written as $\displaystyle (p,q)=(\cos\theta,\sin\theta).$ Then $\displaystyle dp=-\sin\theta\,d\theta, ~ dq=\cos\theta\,d\theta,$ and therefore $\displaystyle \alpha=\cos\theta(\cos\theta\,d\theta)-\sin\theta(-\sin\theta\,d\theta)=d\theta.$ Thus $\alpha$ is the infinitesimal signed-angle form. The angle $\theta$ itself cannot be chosen as a single global real-valued function on the whole circle, but its differential $d\theta$ is globally meaningful, and the formula above defines it without choosing a branch of angle.

There is also a geometric interpretation that will generalize. At $y=(p,q)\in S^1$ , let $v=(a,b)$ be tangent to the circle. Then

$\displaystyle \alpha_y(v)=pb-qa=\det\begin{pmatrix}p&a\\q&b\end{pmatrix}.$

The vector $y$ is the outward unit normal and $v$ is tangential. The determinant is the signed area of their parallelogram; because the normal has length one and is perpendicular to the tangent, it is simply signed tangential length. Positive sign means counterclockwise motion, and negative sign means clockwise motion.

Now let $\displaystyle U:D^2\longrightarrow S^1$ be a smooth map, and write $\displaystyle U(x,y)=(p(x,y),q(x,y)).$ Pull the angle form back to the disk: $\displaystyle \beta:=U^*\alpha=p\,dq-q\,dp.$ The value of $\beta$ on a small displacement in the source disk records the signed angular motion of the target point along the circle.

The central calculation is $\displaystyle d\beta=0.$ Indeed,

$\displaystyle d\beta=d(p\,dq)-d(q\,dp)=dp\wedge dq-dq\wedge dp=2\,dp\wedge dq.$

Writing $\displaystyle dp=p_x\,dx+p_y\,dy,\qquad dq=q_x\,dx+q_y\,dy,$ we obtain

$\displaystyle dp\wedge dq=(p_xq_y-p_yq_x)\,dx\wedge dy.$

The coefficient $p_xq_y-p_yq_x$ is the Jacobian determinant of the map $U$ considered as a map into the ambient plane. But $U$ takes values in the unit circle, so $\displaystyle p^2+q^2=1.$ Differentiating with respect to $x$ and $y$ gives $\displaystyle pp_x+qq_x=0,\quad pp_y+qq_y=0.$ Thus the two derivative vectors $\displaystyle U_x=(p_x,q_x),\quad U_y=(p_y,q_y)$ are both perpendicular to $U=(p,q)$ . In the plane, the perpendicular space to a nonzero vector is one-dimensional. Hence $U_x$ and $U_y$ are parallel, so their determinant vanishes: $\displaystyle p_xq_y-p_yq_x=0.$ Therefore

$\displaystyle dp\wedge dq=0,\quad d\beta=0.$

This is the whole local mechanism in dimension two. A tiny two-dimensional source square has two independent tangent directions, but after applying $U$ both directions lie in the one-dimensional tangent line of the circle. The image of the square therefore has no first-order planar area.

Stokes’ theorem now turns this pointwise statement into a boundary statement. For any small rectangle $Q\subset D^2$ , $\displaystyle \int_{\partial Q}\beta=\int_Qd\beta=0.$ This does not say that the angular change along each edge is zero. It says that the total signed angular change around the four edges is zero. Locally, where one can choose a genuine angle function $\theta$ , one has $\beta=d\theta$ , and the increments telescope around the closed loop.

Tiling the disk by tiny rectangles makes the global cancellation visible. Each interior edge is traversed twice with opposite orientations, so its two contributions cancel. Only the outer boundary remains. Consequently,

$\displaystyle \int_{S^1}U^*\alpha=0.$

Every smooth map from the disk to the circle has zero total signed boundary angle.

By contrast, the identity map of the circle has $\displaystyle \int_{S^1}\alpha=\int_0^{2\pi}d\theta=2\pi.$

Thus the identity map of $S^1$ cannot extend across the disk while remaining circle-valued. This is the analytic obstruction in its simplest form.

Higher dimensions

For $S^1$ , the natural infinitesimal quantity is signed length, or signed angle. For $S^2$ , it is signed surface area. For $S^{n-1}$ , it is signed $(n-1)$ -dimensional area.

Let $\displaystyle \Omega:=dy_1\wedge\cdots\wedge dy_n$ be the standard oriented volume form on $\mathbb R^n$ . Define the $(n-1)$ -form $\displaystyle \alpha:=\sum_{j=1}^n(-1)^{j-1}y_j\,dy_1\wedge\cdots\wedge\widehat{dy_j}\wedge\cdots\wedge dy_n.$ The hat means that the indicated factor is omitted. The coordinate formula becomes transparent when we evaluate it. For $y,v_1,\ldots,v_{n-1}\in\mathbb R^n$ , $\displaystyle \alpha_y(v_1,\ldots,v_{n-1})=\det[y,v_1,\ldots,v_{n-1}].$

Now suppose that $y\in S^{n-1}$ and the vectors $v_1,\ldots,v_{n-1}$ lie in the tangent hyperplane $\displaystyle T_yS^{n-1}=y^\perp.$ The vector $y$ is the outward unit normal. Thus the determinant is the signed $(n-1)$ -dimensional area of the parallelepiped spanned by $v_1,\ldots,v_{n-1}$ . Its magnitude is $\displaystyle \left|\det[y,v_1,\ldots,v_{n-1}]\right|=\sqrt{\det\bigl(\langle v_i,v_j\rangle\bigr)_{1\le i,j\le n-1}},$ the ordinary Riemannian area density. Its sign records whether the ordered tangent frame agrees with the orientation induced by the outward normal. Thus the restriction of $\alpha$ to $S^{n-1}$ is the usual oriented area form. For $S^1$ , this reduces to $\displaystyle \alpha_y(v)=\det[y,v],$ which is signed tangential length. For $S^2$ , $\displaystyle \alpha_y(v,w)=\det[y,v,w].$ The cross product $v\times w$ measures the oriented area of a tiny tangent parallelogram, and the dot product with the outward normal decides its sign. More generally, if $\displaystyle \Phi(s_1,\ldots,s_{n-1})$ parametrizes a patch of $S^{n-1}$ , then

$\displaystyle \Phi^*\alpha=\det\left[\Phi,\frac{\partial\Phi}{\partial s_1},\ldots,\frac{\partial\Phi}{\partial s_{n-1}}\right]ds_1\wedge\cdots\wedge ds_{n-1}.$

Integrating this form counts the target sphere-area swept out by $\Phi$ , with multiplicity and sign. Patches covered with the reversed orientation contribute negatively. This is the direct higher-dimensional analogue of integrating $d\theta$ around the circle. The form $\alpha$ satisfies the elementary identity $\displaystyle d\alpha=n\Omega.$ Indeed, differentiating the $j$ -th summand differentiates only $y_j$ ; after placing the resulting $dy_j$ in its natural position in the wedge product, every term becomes $\Omega$ . There are $n$ such terms.

Now let $\displaystyle H:C\longrightarrow S^{n-1}$ be a smooth map from an $n$ -dimensional region $C$ . In the fixed-point proof, $C$ will be either the ball $B^n$ or the cylinder $S^{n-1}\times[0,1]$ . Since $H$ lands in the sphere, $\displaystyle |H(z)|^2=1.$ Differentiating in any source direction $\xi$ gives $\displaystyle 0=D_\xi\langle H,H\rangle=2\langle H,DH(\xi)\rangle.$ Therefore $\displaystyle DH(\xi)\in H(z)^\perp=T_{H(z)}S^{n-1}.$ Every derivative vector lies in an $(n-1)$ -dimensional tangent hyperplane. If $\xi_1,\ldots,\xi_n$ are $n$ source tangent vectors, then the $n$ vectors $\displaystyle DH(\xi_1),\ldots,DH(\xi_n)$ are linearly dependent. Hence

$\displaystyle \det\bigl[DH(\xi_1),\ldots,DH(\xi_n)\bigr]=0.$

But this determinant is exactly the pullback of the ambient volume form:

$\displaystyle (H^*\Omega)(\xi_1,\ldots,\xi_n)=\Omega\bigl(DH(\xi_1),\ldots,DH(\xi_n)\bigr).$

Thus $\displaystyle H^*\Omega=0.$ A tiny $n$ -dimensional source cell may be folded and stretched by $H$ , but to first order it is forced into an $(n-1)$ -dimensional tangent plane of the sphere. It has no genuine $n$ -dimensional target volume.

Using the chain rule for differential forms, we obtain $\displaystyle d(H^*\alpha)=H^(d\alpha)=nH^*\Omega=0.$ This is the local conservation law. The form $H^*\alpha$ measures signed sphere-area carried by infinitesimal $(n-1)$ -faces in the source, and its exterior derivative says that the signed total carried by the faces of every infinitesimal $n$ -cell is zero.

For a small $n$ -cell $Q$ , Stokes’ theorem gives

$\displaystyle \int_{\partial Q}H^*\alpha=\int_Qd(H^*\alpha)=0.$

When neighboring cells are added together, each common interior face occurs twice with opposite orientations, and the two contributions cancel. Only the outer boundary survives. In particular, if $\displaystyle U:B^n\longrightarrow S^{n-1}$ is smooth, then

$\displaystyle \int_{S^{n-1}}U^*\alpha=\int_{\partial B^n}U^*\alpha=\int_{B^n}d(U^*\alpha)=0.$

And if $\displaystyle H:S^{n-1}\times[0,1]\longrightarrow S^{n-1}$ is a smooth sphere-valued homotopy, then the boundary of the cylinder is the top sphere minus the bottom sphere. Hence

$\displaystyle 0=\int_{\partial(S^{n-1}\times[0,1])}H^*\alpha=\int_{S^{n-1}}H_1^*\alpha-\int_{S^{n-1}}H_0^*\alpha.$

Therefore $\displaystyle \int_{S^{n-1}}H_1^*\alpha=\int_{S^{n-1}}H_0^*\alpha.$ So total signed sphere-area is unchanged under a sphere-valued deformation. This is the analytic content normally summarized by the word “degree,” but no separate theory of degree is required here: the statement follows from rank deficiency and Stokes’ theorem.

Return to the fixed-point-free map $\displaystyle f:B^n\longrightarrow B^n.$ Its normalized displacement field $\displaystyle u(x)=\frac{x-f(x)}{|x-f(x)|}$ is defined on all of $B^n$ , so the extension calculation gives

$\displaystyle \int_{S^{n-1}}u^*\alpha=0.$

On the other hand, on the boundary we have the homotopy $\displaystyle H(x,t)=\frac{x-tf(x)}{|x-tf(x)|},$ from the identity map to $u|_{S^{n-1}}$ . Homotopy invariance therefore gives

$\displaystyle \int_{S^{n-1}}u^*\alpha=\int_{S^{n-1}}{\text{id}}^*\alpha.$

Since ${\text{id}}^*\alpha=\alpha$ and $\alpha$ is the oriented area form of the unit sphere,

$\displaystyle \int_{S^{n-1}}{\text{id}}^*\alpha=\int_{S^{n-1}}\alpha={\text{Area}}(S^{n-1})>0.$

Thus the same quantity is both zero and strictly positive:

$\displaystyle 0=\int_{S^{n-1}}u^*\alpha={\text{Area}}(S^{n-1}).$

This contradiction shows that a smooth self-map of the ball must have a fixed point.

The underlying picture is simple. On the boundary, the direction $x-f(x)$ can be deformed into the outward normal $x$ . The outward normal covers the target sphere once positively: on the circle it makes one positive full turn, on the two-sphere it sweeps every oriented area patch once, and in general it has one net positive signed coverage. But a sphere-valued field that extends through the whole ball has zero net signed boundary coverage, because every tiny interior cell contributes zero net boundary area and all interior faces cancel.

Smooth to Continuous

Now let $\displaystyle f:B^n\longrightarrow B^n$ be just continuous and suppose it has no fixed point. By compactness, $\displaystyle \delta:=\min_{x\in B^n}|x-f(x)|>0.$ Choose $\varepsilon>0$ with $3\varepsilon<\delta$ , and set $\displaystyle f_0=(1-\varepsilon)f.$ Then $f_0(B^n)$ lies in the smaller ball of radius $1-\varepsilon$ , while $\displaystyle |x-f_0(x)|\ge\delta-\varepsilon>0.$ Extend $f_0$ to $\mathbb R^n$ by $\displaystyle \widetilde f_0(x)=f_0\left(x/\max{1,|x|}\right),$ and convolve with a nonnegative mollifier $\rho_\eta$ . For sufficiently small $\eta$ , the smooth map

$\displaystyle g_\eta(x):=\int_{\mathbb R^n}\rho_\eta(z)\widetilde f_0(x-z),dz$

satisfies $\displaystyle \sup_{B^n}|g_\eta-f_0|<\varepsilon.$ Since $g_\eta(x)$ is an average of points in the convex ball of radius $1-\varepsilon$ , it remains a smooth self-map of $B^n$ ; moreover,

$\displaystyle |x-g_\eta(x)|\ge|x-f(x)|-|f-f_0|-|f_0-g_\eta|>\delta-2\varepsilon>0.$

Thus $g_\eta$ is a smooth fixed-point-free self-map of the ball, contradicting the smooth case. Hence every continuous map $\displaystyle f:B^n\longrightarrow B^n$ has a fixed point.

Conclusion:

The proof is driven by one local geometric fact:

$\displaystyle \text{a map into }S^{n-1}\text{ has no infinitesimal }n\text{-dimensional volume}.$

For a disk-to-circle map, this becomes the assertion that no infinitesimal two-dimensional area is produced, and hence that the total boundary angle is zero. For maps into higher spheres, it becomes vanishing ambient volume and hence zero total signed boundary sphere-area. Stokes’ theorem is the device that assembles this infinitesimal vanishing into a global obstruction. The fixed-point theorem is therefore not a mysterious global miracle. A fixed-point-free map would construct a sphere-valued direction field on the ball whose boundary behavior is forced to resemble the outward normal. But the outward normal has nonzero total signed coverage of the sphere, whereas every sphere-valued field extending over the ball has zero total signed coverage. The two requirements are incompatible.

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