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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.

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  • $\begingroup$ If $f^{-1}$ exists, then $f$ is bijective, so the condition of injection is redundant. $\endgroup$
    – Yes
    Oct 5, 2015 at 3:14
  • $\begingroup$ @Gudson Chou The condition of injectivity is redundant, but decidedly not for the reasons you claim. $f^{-1}$ always exists as a partial multi-valued function, and as a partial function (so that we can make sense of "increasing") for one-to-one $f$. The condition is redundant, however, because increasing functions are always one-to-one. $\endgroup$
    – Marcel Besixdouze
    Oct 5, 2015 at 4:58
  • $\begingroup$ @MarcelT.: That is a generalization. I do not believe the question assigned to the OP considers partial function. Moreover, unless otherwise stated, people do not consider partial functions like you do not tell people that which addition and multiplication you are using. Thus, I do not see you have a benign motive to speak of partial functions in the present context. $\endgroup$
    – Yes
    Oct 5, 2015 at 5:05
  • $\begingroup$ Primary school algebra will define $f^{-1}$ as the function with domain the range of $f$ and whose graph is obtained from $y=f(x)$ by interchanging $x$ and $y$, if $f$ passes the horizontal line test. This is defined far before bijectivity. There is no reason to imagine when the OP wrote "one-to-one" the OP meant "bijective". The statement does not require bijectivity at all. It doesn't even name spaces for the concept of surjectivity to even make any sense at all. "Partial function" is not a generalization, students are taught early that real-valued functions can have restricted domains. $\endgroup$
    – Marcel Besixdouze
    Oct 5, 2015 at 5:34
  • $\begingroup$ @MarcelT.: The usual definition is that if a function is not bijective then the inverse is not defined, lest you be unaware. $\endgroup$
    – Yes
    Oct 5, 2015 at 6:16

5 Answers 5

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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.

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$f(f^{-1}(x))=x$

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 $f^{-1}(x)$?

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    $\begingroup$ This is just restating the question. -1 $\endgroup$
    – Marc van Leeuwen
    Oct 5, 2015 at 6:02
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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 $u_0,v_0$.

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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 $y=f^{-1}(b)$.

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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.

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