I've been reading this article that proposes an inductive proof of Brouwer's fixed point theorem just using "basic" topology, avoiding the usage of homotopy groups as the usual proof goes.
To do so we assume that given $n \in \mathbb{N}$ any continuous function from $C = [0,1]^n$ to $C$ has a fixed point and we prove that any continuous function $f$ from $C\times[0,1]$ to $C \times [0,1]$ has a fixed point (the base case for $n = 1$ is already proven by the intermediate value theorem). We write $f = (f_1,\dots,f_{n+1})$ where each $f_i:C\times [0,1] \rightarrow [0,1]$ is a continous function. Next it defines $\phi = (f_1,\dots,f_n)$ which is clearly a continuous function and we write $\phi_t: C \rightarrow C$ as the function $\phi_t(x) = \phi(x,t)$, which clearly is continuous. Next by induction hypothesis we note that each $\phi_t$ has a fixed point, with this in mind we define $$X = \{(x,t) \in C \times [0,1]: \phi_t(x) = x\} = (\phi-id)^{-1}(\{0\})$$
And because this set is the preimage of a closed set under a continuous function, it's closed and nonempty because by induction each $\phi_t$ has at least one fixed point.
However here's where my problem with the proof comes because next it takes $\pi_2: C\times [0,1] \rightarrow [0,1]$ as the projection onto the last coordinate. We note that $\pi_2(X) = [0,1]$ as there's at least one fixed point for each $\phi_t$, but the problem that because clearly $\pi_2$ is an open map the proof on the article asumes that $\pi_2|_X$ is also an open map. However, restrictions of open maps are not always open, so is the article wrong or what am I missing?
I've tried to prove that this map is in fact open, but I can't seem to do so, and in fact it looks that it might not be open which would prove that the article is wrong, so how do you prove that in fact this restriction is open? Or what would be a counterexample of a function $f$ such that the set we defined as $X$ is such that the restriction of $\pi_2$ isn't open?
Best Answer
The claim that $\pi_2$ restricted to $X$ is open is unfortunately false. Consider
$$f:[0,1]^2\to[0,1]^2$$ $$f(x,t)=(tx,t)$$
Then $\phi(x,t)=tx$ and $\phi_t(x)=tx$. Therefore $\phi_t(x)=x$ if and only if $x=0$ or $t=1$. In other words
$$X=\big(\{0\}\times[0,1]\big)\cup\big([0,1]\times\{1\}\big)$$
You can verify very easily that neither of the projections restricted to $X$ is open (some of its open subsets are mapped onto a point).
After reading the article this looks like a crucial, unrecoverable mistake in the proof. Or at least I don't see how this can be fixed. I might be wrong though. Or maybe there is no elementary proof after all...