Brouwer Fixed Point Theorem for the Disk

From a Fixed-Point-Free Map to a Retraction

The disk and its boundary are closely linked: if a point in the disk is moved to a different point of the disk, the ray from its image through the original point has a further intersection with the boundary. If this construction can be made continuously for every point, it would give a retraction of the disk onto its boundary. The no-retraction theorem rules that out, and therefore rules out a fixed-point-free map.

We use the Euclidean topology on \(\mathbb{R}^2\) and the subspace topologies on the closed unit disk and its boundary circle. Write points as vectors in \(\mathbb{R}^2\), use \(\cdot\) for the Euclidean dot product, and use \(\|\,\cdot\,\|\) for the Euclidean norm.

The two degree facts in this statement apply to the same boundary map: \(r|_{S^1}=\operatorname{id}_{S^1}\), because \(r\) fixes every boundary point. Since \(r\) itself extends this boundary map over the disk, the extension result gives degree \(0\); since the boundary map is the identity, its degree is \(1\). The contradiction \(0\ne1\) proves the no-retraction theorem.

The Boundary Intersection Formula

Suppose, toward a contradiction, that \(f:D^2\to D^2\) is continuous and has no fixed point. For each \(x\in D^2\), put \(y=f(x)\) and \(v=x-y\). The assumption that \(f(x)\ne x\) ensures \(v\ne0\). The ray starting at \(y\) and travelling in direction \(v\) is parametrized by \(y+tv\), where \(t\ge0\); the point \(x\) occurs at \(t=1\).

To find where this ray meets the circle, solve \(\|y+tv\|^2=1\). Expanding the dot product gives a quadratic equation in \(t\):

Because \(\|v\|^2>0\), the larger root is defined by the quadratic formula. Use that root to define \(R(x)\):

We check that the formula really selects the boundary point farther along the ray. The quadratic \(q(t)=\|y+tv\|^2-1\) has positive leading coefficient. Since \(y=f(x)\in D^2\) and \(x\in D^2\), we have \(q(0)=\|y\|^2-1\le0\) and \(q(1)=\|x\|^2-1\le0\). Thus \(t=1\) lies between the two roots, allowing equality at a root. In particular, the larger root \(t_+(x)\) is at least \(1\). Substitution into the quadratic gives \(\|R(x)\|=1\), so \(R(x)\in S^1\).

The formula is continuous in \(x\). Indeed, \(f\) is continuous; therefore \(y=f(x)\), \(v=x-f(x)\), and the dot products and norms in the formula vary continuously with \(x\). The denominator \(\|v\|^2\) never vanishes by the fixed-point-free assumption. The expression under the square root is nonnegative: it is the discriminant of the quadratic, which has real roots because \(q(t)=0\) has a solution at the boundary intersection of the ray. Equivalently, the displayed formula itself gives the nonnegative square root of that discriminant. Hence \(t_+\), and then \(R\), are continuous.

It remains to check that \(R\) fixes the boundary. If \(x\in S^1\), then \(q(1)=0\). If \(q(0)<0\), the two roots have opposite signs, so the positive root is \(1\), and it is the larger root. If \(q(0)=0\), then \(y\in S^1\); the roots are \(0\) and \(1\), because \(q(1)=0\) and the quadratic has nonzero leading coefficient. Again the larger root is \(1\). Thus \(t_+(x)=1\) in either case, and \(R(x)=y+v=x\). We have shown that \(R:D^2\to S^1\) is continuous and fixes every point of \(S^1\), so it is a retraction.

Brouwer’s Fixed Point Theorem for the Disk

Proof. Assume for contradiction that \(f:D^2\to D^2\) is continuous and has no fixed point. The boundary intersection construction above then produces a continuous retraction \(R:D^2\to S^1\). The no-retraction theorem says that no such map exists. Therefore the assumption that \(f\) has no fixed point is false, and there is \(x\in D^2\) with \(f(x)=x\). \(\square\)

The argument requires both hypotheses on \(f\). Continuity is what makes the constructed boundary point vary continuously, and the condition \(f(D^2)\subseteq D^2\) ensures the ray starts at a point of the disk and that its quadratic has the required boundary intersection. Without a self-map of the disk, the conclusion need not hold.

A Common Pitfall and Why the Argument Matters

Common pitfall. One might try the map \(f(x)=-x\) as a fixed-point-free example and then use the ray formula. But the fixed-point-free assumption fails at the origin: \(f(0)=0\). At that point \(v=x-f(x)=0\), so the denominator \(\|v\|^2\) in the formula is zero. Dividing by it would be invalid, not a way to define the missing boundary point. In fact, solving \(f(x)=x\) gives \(-x=x\), so \(2x=0\) and \(x=0\); this map has a fixed point, exactly as the theorem requires.

The essential logic is not that every ray construction automatically defines a retraction. It works only under the supposition that \(f(x)\ne x\) at every point, which keeps the direction vector nonzero and makes the formula continuous across the whole disk. Once that condition is imposed, the constructed map contradicts the no-retraction theorem. The contradiction establishes existence of a fixed point without needing to identify where it is for an arbitrary \(f\).

This result is a basic example of a topological obstruction: a proposed map is ruled out because it would force another map with a forbidden property. Here the degree of the boundary identity prevents retraction, while the ray construction shows that a fixed-point-free self-map would force one. The disk theorem is the two-dimensional case of a broader family of fixed-point results, but this argument uses only the disk, its circle boundary, and the no-retraction theorem.