# Why is this easy “proof” of Brouwer's Fixed Point Theorem not correct/common?

Brouwer's Fixed Point Theorem states, essentially, that any continuous function on a closed disc to itself has a fixed point. I am familiar with the proof based on the impossibility of a retraction from a disc to its boundary and the proof based on Sperner's Lemma. Wikipedia lists a number of other proofs.

However, it seems there is a simpler "proof" - quoted because it could be wrong - that uses no fancy machinery and I'm wondering if it is right and if so, why it isn't well known.

We will prove that any continuous $f : [0, 1]^n \rightarrow [0, 1]^n$ has a fixed point by induction on $n$. $n = 1$ amounts to the Intermediate Value Theorem. For $n > 1$ our space is $[0, 1] \times [0, 1]^{n-1}$. By the 1-d case, for each $\mathbf{u} \in [0, 1]^{n-1}$, $x \mapsto f(x, \mathbf{u})_1$, the first component of $f(x, \mathbf{u})$, has a fixed point $x$. By continuity of $f$ we may choose this fixed point, $x(\mathbf{u})$, to vary continuously in $\mathbf{u}$. By the $(n-1)$-d case, for each $y \in [0, 1]$, $\mathbf{v} \mapsto f(y, \mathbf{v})_{2, \ldots, n}$ has a fixed point $\mathbf{v}$ and we may let $\mathbf{v}(y)$ vary continuously.

If $x(\mathbf{v}(0)) = 0$ then $(0, \mathbf{v}(0))$ is a fixed point of $f$; similarly for 1. Otherwise let $X = \{(x(\mathbf{u}), \mathbf{u}) \mid u \in [0, 1]^{n-1}\}$. $X$ is the graph of a continuous function so it is closed and $[0, 1]^n \setminus X$ is open. Furthermore, $\mathbf{v}(0)$ and $\mathbf{v}(1)$ are in different components of $[0, 1]^n \setminus X$ so by an argument like the proof of the Intermediate Value Theorem, $\mathbb{v}$ must cross $X$ at some point, which is a fixed point of $f$.

• I think the reason the textbooks exclude this is the same reason you include it: it doesn't use any other machinery (except calculus). I'm sure some authors have thought of this because I also independently came up with something similar. – Jacob Wakem Jul 8 '16 at 15:33
• Showing that $x(u)$ varies continuously in $u$ seems to be nontrivial. There might be more than one fixed point for each $u$. – Seewoo Lee Jul 8 '16 at 15:35
• @See-WooLee You are right, that seems to be the catch. – Solomonoff's Secret Jul 8 '16 at 15:45
• I'm satisfied that you've reduced the proof to a proof of the $[0,1]^2 \to [0,1]^2$ case (assuming that the fixed point can be continuously chosen). However, I don't quite see that you've satisfactorily proved a fixed point in this case. – Omnomnomnom Jul 8 '16 at 15:52
• "By continuity, we can pick..." I don't see how you can be sure to do that. That's a much bigger leap than you seem to think it is. – Thomas Andrews Jul 8 '16 at 16:32

In general, $x(\mathbf{u})$ can't be chosen to be continuous. Here is a simple counter-example in $\mathbb R^2$:
$$f(x,y) = \begin{cases} 2xy, & \text{if } y \le \frac12 \\ 1-2(1-x)(1-y), & \text{if } y \ge \frac12 \end{cases}$$
If $y < \frac12$, the unique fixed $x$ is $x=0$; if $y > \frac12$, the unique fixed $x$ is $x=1$. (If $y=\frac12$, then $f(x,y)=x$ fixes all points in $[0,1]$.)
I think your proof works fine - as long as the fixed points involved are all unique. What if the map taking $x$ to $(x, \mathbf {u} )$ has multiple fixed points, so your $x (\mathbf {u})$ is ill-defined? There might not be a continuous curve of fixed points over the whole space.
• Note that the proof arbitrarily chooses some continuous $x(\mathbb{u})$. There may be additional fixed points not chosen. – Solomonoff's Secret Jul 8 '16 at 16:07
• What I mean is there might be no such curve. For example, in the case of $[0,1]^2$, the fixed points could make two disjoint parabolas that together cover all values of each coordinate. – Reese Jul 8 '16 at 16:45