Prove that if $f$ is increasing then so is $f^{-1}$ Prove that if $f$ is increasing then so is $f^{-1}$, when $f$ is a one-to-one function.
I'm having trouble figuring out how to get started with this question. I'm assuming it has something to do with the definition of infectivity, but that's as far as i've gotten.
 A: Let $f\colon A\to B$ an increasing and bijective function.
Suppose $f^{-1}$ is not increasing. There exists $y_1$ and $y_2$ in $B$ such that $y_1 < y_2$ and $f^{-1}(y_1) \geqslant f^{-1}(y_2)$. As $f$ is increasing, we have $f(f^{-1}(y_1)) \geqslant f(f^{-1}(y_2))$, that is $y_1\geqslant y_2$. Contradiction: the function $f^{-1}$ is increasing.
A: $f(f^{-1}(x))=x$ 
The right hand side is always increasing in x so the left hand side must also be increasing in x. You know that $f(\cdot)$ is increasing. What does that mean for $f^{-1}(x)$?
A: A non-derivative proof.
What we have is $\forall u,v. (u>v\implies f(u)>f(v))$.
We want to prove $\forall u,v.  (f(u)>f(v)\implies  u>v)$.
Suppose the opposite, that there exist $u_0,v_0$ such that $f(u_0)>f(v_0)$ is true and $u_0>v_0$ is false for the sake of contradiction.
Since $u_0>v_0$ is false, $u_0=v_0$ or $u_0<v_0$.  
Case 1. $u_0=v_0$. 
Then $f(u_0)=f(v_0)$, contradiction.
Case 2. $u_0 < v_0$. Then since $\forall v,u. (v>u\implies f(v)>f(u))$, we have $f(v_0)>f(u_0)$, contradiction.
Hence there cannot exist such $u_0,v_0$.
A: That $f$ is increasing means that $x\leq y\to f(x)\leq f(y)$ holds. Then  also $x<y\to f(x)<f(y)$ since $f$ is injective, as well as $f(y)<f(x)\to y<x$ by contrapositive, which is the same as $f(x)<f(y)\to x<y$ (only the names of the variables were interchanged). So $f$ increasing means $x<y\leftrightarrow f(x)<f(y)$.
Now if $f$ has inverse $f^{-1}$, one has
$$
  \hbox{$f$ increasing}\iff x<y\leftrightarrow f(x)<f(y) \implies
  f^{-1}(a)<f^{-1}(b)\leftrightarrow f(f^{-1}(a))<f(f^{-1}(b)) \\
  \iff f^{-1}(a)<f^{-1}(b)\leftrightarrow a<b \iff \hbox{$f^{-1}$ increasing},
$$
where the implication step specialises $x=f^{-1}(a)$ and $y=f^{-1}(b)$.
A: Suppose by way of contradiction that $f^{-1}$ were not increasing. 
That is, suppose that there were $y_1<y_2$ in the range of $f$ for which $f^{-1}(y_1)\ge f^{-1}(y_2)$. Applying $f$, we get $f(f^{-1}(y_1))\ge f(f^{-1}(y_2))$. Cancelling out $f$ and $f^{-1}$ we get $y_1 \ge y_2$, a contradiction to our assumption that $y_1<y_2$. 
So $f^{-1}$ is increasing. 
