Fixed Point Property for a special space? Suppose $X$ is a compact connected metric connected space and for every $\epsilon >0$ , there exists a continuous surjective function $f : X \rightarrow [0,1]$, that for all $y \in [0,1]$, the diameter of the set $f^{-1} (y)$ is less than $\epsilon$.
Prove that $X$ has the fixed point property; that is, for all continuous $g : X \rightarrow X$, there exists $x_0 \in X$ that $g(x_0) = x_0$.
If we know that for a continuous surjective function $f : X \rightarrow [0,1]$ the diameter of set $f^{-1} (y)$ is zero, then we know that $X$ and $[0,1]$ are homeomorphic, so the statement is proven. But I have no idea for that.
Hints are appreciated.
 A: Here is one solution which works even when you replace the closed unit interval with the closed unit $N$-ball $B$. I will describe steps of the proof and leave you to fill in the details. (Since the problem smells like a homework.) 
(0). Prove the following lemma akin to the uniform continuity of continuous functions on compact metric spaces (and proven in a similar fashion):
Lemma. Let $f: (X,d_X)\to (Y,d_Y)$ be a continuous map between two compact metric spaces such that for every $y\in Y$, $diam(f^{-1}(y))\le \epsilon$. Then there exists $\sigma>0$ (depending on $f$ and on $\epsilon$) such that 
$$
\forall x_1, x_2\in X, ~~~d_Y(f(x_1), f(x_2))<\sigma \Rightarrow d_X(x_1, x_2)< 2\epsilon. 
$$
Proof. Suppose the claim fails. Then for every natural number $n$ there exists a pair of points $x_n, x'_n\in X$ such that $d_X(x_n, x'_n)\ge
 2\epsilon$ while 
$$
d_Y(f(x_n), f(x'_n))<\frac{1}{n}.$$
In view of compactness of $X$, after passing to a subsequence, we can assume that 
$$
\lim_{n\to\infty} x_n=x, \lim_{n\to\infty} x'_n=x',
$$
with $d_X(x,x')\ge 2\epsilon$, while
$$
\lim_{n\to\infty} f(x_n)= \lim_{n\to\infty} f(x'_n)=y\in Y,
$$
and $f(x)=f(x')=y$. Thus, the preimage of $y$ has diameter $\ge 2\epsilon$. A contradiction. qed
(1). Given $f$ and $g$ as in your problem ($f$ is surjective and $diam(f^{-1}(y))<\epsilon$ for all $y\in B$), construct a piecewise-linear continuous map $h: B\to B$ such that the following diagram is "almost commutative":
$$
\require{AMScd}
\begin{CD}
X @>{g}>> X\\
@VVfV @VVfV \\
B @>{h}>> B
\end{CD}
$$
meaning that $d(f\circ g, h\circ f)<\delta(\epsilon)$, where
$$
\lim_{\epsilon\to 0}\delta(\epsilon)=0. 
$$
I will explain how to do this in the case $B=[0,1]$, the extension to the case when $B$ is higher-dimensional is quite straightforward. 
In order to construct such $h$ first pick a finite subset $0=y_0< y_1 < y_2 < ... <y_n\in [0,1]$ (with $|y_i- y_{i-1}|$ sufficiently small for all $i$) such that 
$$
\forall i\in \{1,...,n\}, ~~diam(f^{-1}([y_{i-1}, y_i])) < 2\epsilon 
$$
Then define $h$ on the finite subset $\{y_0,...,y_n\}$ so that 
$$
\forall y_i, \exists x_i\in f^{-1}(y_i), h(y_i)=fg(x_i).
$$
Then extend $h$ to the rest of $[0,1]$ linearly on each interval $[y_{i-1},y_i]$. 
(2). Use the fact that $B$ has the fixed point property to show that $g$ "almost" has a fixed point, i.e.: For every $\epsilon>0$ there is a point $x\in X$ such that $d(g(x), x)\le \eta(\epsilon)$, where 
$$
\lim_{\epsilon\to 0}\eta(\epsilon)=0. 
$$
To prove this, take $y\in B$ such that $h(y)=y$ and think of its preimage $f^{-1}(y)$. 
(3). Conclude that there is a sequence $(x_n)$ in $X$ such that
$$
\lim_{n\to\infty} d(g(x_n), x_n)=0.
$$
(4). Use compactness of $X$ to show that $g$ has a fixed point in $X$. 
A: consider $h: X \rightarrow X$ continuous. I want to prove there's $x \in X$ that $h(x)=x$. for $\epsilon > 0$ consider $foh -f : X \rightarrow [0,1]$.because $f$ is surjective, there is $x \in X$  that $(foh -f) (x) \leq 0$ and there is $y \in X$ that $(foh -f)  (y)\geq 0$. $foh -f $ is continuous and $X$ is connected, so there exists $t \in X$ that $(foh -f)(t) =0$. then $t , h(t) \in f^{-1}(f(t))$. so $d(t , h(t)) < \epsilon$.so we can have a sequence $t_n$ that $d(t_n , h(t_n)) < 1/n$. because $X$ is compact we can find $s \in X$ that $s = h(s)$.
