Prove that, for every continuous map \(f : B^2 \to B^2\), there is a point \(x\) such that \(f (x) = x\).
This is the \(n = 2\) case of the Brouwer fixed point theorem; your proof shouldn’t appeal to either of the Brouwer or the Lefschetz fixed point theorems.
Solution
1. Suppose, for contradiction, that\[ f:B^2\to B^2 \]has no fixed point.Then\[ g:B^2\to S^1, \qquad g(x)=\frac{x-f(x)}{\|x-f(x)\|}, \]is a well-defined continuous map.
The assumption that \(f\) has no fixed point says \[ x-f(x)\ne0 \] for every \(x\in B^2\). Thus the denominator never vanishes. The numerator is continuous, normalization is continuous on \(\mathbb R^2\setminus\{0\}\), and the normalized vector has norm \(1\). Hence \(g\) is a continuous map into \(S^1\).
2. On the boundary \(S^1=\partial B^2\), the map \(g|_{S^1}\) is homotopic to the identity map of \(S^1\).
For \(x\in S^1\) and \(0\le t\le1\), set \[ H(x,t)=\frac{x-tf(x)}{\|x-tf(x)\|}. \] We first show that the denominator never vanishes. If \[ x=tf(x), \] then, because \(\|x\|=1\) and \(\|f(x)\|\le1\), \[ 1=t\|f(x)\|\le t\le1. \] Hence equality holds throughout, so \[ t=1 \qquad\text{and}\qquad f(x)=x, \] contradicting the fixed-point-free assumption.
Therefore \(H\) is well-defined and continuous. At the endpoints, \[ H(x,0)=x \] and \[ H(x,1) = \frac{x-f(x)}{\|x-f(x)\|} = g(x). \] Thus \[ g|_{S^1}\simeq\operatorname{id}_{S^1}. \]
3. Consequently,\[ (g|_{S^1})_*:\pi_1(S^1)\to\pi_1(S^1) \]is the identity homomorphism.
Homotopic maps induce the same homomorphism on fundamental groups, up to the standard basepoint-change isomorphism when the homotopy moves the basepoint. Since \(\pi_1(S^1)\cong\mathbb Z\) is abelian, this basepoint change has no effect on the induced endomorphism. By 2, \[ g|_{S^1}\simeq\operatorname{id}_{S^1}, \] so \[ (g|_{S^1})_*=\operatorname{id}_{\mathbb Z}. \] In particular it is nonzero.
4. But because \(g|_{S^1}\) extends over \(B^2\), its induced map on \(\pi_1\) must be zero.
Let \[ i:S^1\hookrightarrow B^2 \] be the inclusion. Then \[ g|_{S^1}=g\circ i. \] Functoriality of the fundamental group gives \[ (g|_{S^1})_*=g_*\circ i_*. \] Since \(B^2\) is contractible, \[ \pi_1(B^2)=0. \] Therefore \[ i_*:\pi_1(S^1)\to\pi_1(B^2) \] is the zero homomorphism, and hence \[ (g|_{S^1})_*=0. \]
5. The contradiction proves that every continuous map \(f:B^2\to B^2\) has a fixed point.
Step 3 gives \[ (g|_{S^1})_*=\operatorname{id}_{\mathbb Z}, \] while 4 gives \[ (g|_{S^1})_*=0. \] These cannot both hold. Thus the assumption in 1 that \(f\) is fixed-point-free is false. Hence there exists \[ x\in B^2 \] such that \[ \boxed{f(x)=x}. \]