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

Question

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 $f$.

Answer 1

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 $[0,1]$.)

Answer 2

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.