Fixed point in one-to-one function on $\mathbb{R}$ Given that $f:[a,b]\to[a,b]$ is a real continuous one-to-one function, and that neither $a$ or $b$ are fixed points of $f$, show that there exists a fixed point in $(a,b)$.


  Proof:


Since $f(a)\ne a$ and $f(b)\ne b$, and $f$ is one-to-one on $[a,b]$, by IVT, $\exists c_1, c_2 \in (a,b)$ such that $f(c_1)=a$ and $f(c_2)=b$. Since $a$ and $b$ are boundary points, $\exists d_1, d_2\in (a,b)$, with $d_1\ne d_2$ (WLOG, let $d_1 &lt; d_2$), such that $f'(d_1)=f'(d_2)=0$. Because of this and since $f$ is continuous, $\exists k\in (d_1,d_2)$ such that $\left| f'(k)\right|=1$. This implies that $f(x)$ coincides with $g(x)=x$ at least once, which implies that a fixed point exists in $(a,b)$.

I think that rigour suffers someplace in this proof. Would appreciate some feedback and suggestions.
I don't really understand your point about $d_1, d_2$, but I'd say $f(x) = a+b-x$ is a counterexample. Plus, you don't even know if $f$ is differentiable. 

Consider now $g(x) =f(x) - x$. $f$ is a continuous bijection, so it's monotone, and you can then easily check that $f(a) =b$ and consequently $f(b) =a$, so $g(a) &gt; 0, g(b) &lt; 0$ and you just need to apply Bolzano.
I don't really understand your point about $d_1, d_2$, but I'd say $f(x) = a+b-x$ is a counterexample. Plus, you don't even know if $f$ is differentiable. 

Consider now $g(x) =f(x) - x$. $f$ is a continuous bijection, so it's monotone, and you can then easily check that $f(a) =b$ and consequently $f(b) =a$, so $g(a) &gt; 0, g(b) &lt; 0$ and you just need to apply Bolzano.