I have tried to answer the following question from my textbook:
Suppose the following function $f: [a,b] \rightarrow [c,d]$ with $a,b,c,d\in\mathbb{R}$ and $a<b$, $c<d$. The function is monotonically increasing and surjective.
a) Proof that f(a)=c and f(b)=d
b) Proof that f is continuous
(Note: later on we will prove that from (a) and (b) follows that f is surjective)
c) Show that the inverse function of f exists and that it is continuous on [c,d]
What I did up until now:
a)
Suppose $f(a)\neq c$ Then $f(a)>c$ because the function is monotonic. Thus there exists a $x \in [a,b]$ such that $f(x)=c$ and $x \neq a$. Thus $f(x)<f(a)$ and thus, because $f$ is monotonically increasing, $x<a$. But then $ x \notin [a,b]$ which is a contradiction.
Proving that $f(b)=d$ goes similarly
b) My teacher showed the following proof, which I don't completly understand.
Choose $x_{1,2}=f^{-1}(f(x)\pm \epsilon)$ (exists because $f$ is surjective)
We know that $x_1 \leq x \leq x_2$. Choose $\delta:=\frac{1}{2}Min(x-x_1,x_2-x)$.
Take $y \in B(x;\delta)$ then $x_1 < y <x_2$ Thus $f(x_1)<f(y)<f(x_2)$, thus $f(x) \in B(f(x),\epsilon)$.
I would like if people could help me by checking my answer with question a and explain why the answer for b is correct. I don't understand how we know that we can take the inverse function of $f$. Could someone explain to me why this is true. I don't know how to write down an answer for question c, it seems logical but I can't explain it.
Best Answer
a) Actually, if $f(a)\neq c$, then $f(a)>c$ simply because the counterdomain of $f$ is $[c,b]$, not because $f$ is monotonic. Also, there is no need for $x\neq c$. You're correct in implying that $f(x)<f(a)$ and thus $x<a$, which is a contradiction.
b) I'm going to do some more details: Let $x\in[a,b]$. For simplicity, suppose that $x\neq a$, $x\neq b$ (otherwise, you have to check only left or right continuity, which is very similar to this case). Let $\varepsilon>0$, and consider $\varepsilon'=\frac{1}{2}\min(d-f(x),f(x)-c,\varepsilon)$ (notice that $\varepsilon'$ depends only on $\varepsilon$), so that $\varepsilon'>0$ and the interval $(f(x)-\varepsilon',f(x)+\varepsilon')$ is contained in $(c,d)$. By surjectivity, there exists $x_{\varepsilon}^-,x_{\varepsilon}^+\in[a,b]$ such that $f(x_{\varepsilon}^\pm)=f(x)\pm\varepsilon'$. Since $f$ is monotonic, then $x_{\varepsilon}^-<x<x_{\varepsilon}^+$.
Let $\delta=\frac{1}{2}\min(x_{\varepsilon}^+-x,x-x_{\varepsilon}^-)$ (notice that $\delta$ depends solely on $\varepsilon$). Given any $y\in(x-\delta,x+\delta)$, we have $x_{\varepsilon}^-<x-\delta<y<x+\delta<x_{\varepsilon}^+$, so, since $f$ is monotonic, $$f(x)-\varepsilon<f(x)-\varepsilon'=f(x_\varepsilon^-)<f(y)<f(x_\varepsilon^+)=f(x)+\varepsilon'<f(x)+\varepsilon$$ because $\varepsilon'<\varepsilon$. Thus, we have just shown that $f(y)\in(f(x)-\varepsilon,f(x)+\varepsilon)$ whenever $y\in(x-\delta,x+\delta)$, so $f$ is continuous at $x$.
Since $x$ was arbitrary (and assuming you have already shown the cases $x=a$ and $x=b$), then $f$ is continuous.
c)By assumption, $f$ is surjective and, since $f$ is monotonic, it is injective, so there exists an inverse $f^{-1}$. This inverse is also surjective and, you can easily check, monotonic. By item b), with $f^{-1}$ in place of $f$, $f^{-1}$ is continuous.