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Let $I,J$ be two interval(which is connected subset on $\Bbb{R}$).Let $f:I\to J$ be diffeomorphism .Prove $f$ is stricly monotone.

My attempt:

Let two endpoint of $I$ be $x_0,x_1$ ($x_0 < x_1$) (If no endpoint we may extend to it by continuity),$f(x_0) = y_0$ and $f(x_1) = y_1$.Wlog we may assume $y_0<y_1$. (then the problem becomes to prove that this map is strictly monotone increasing)

Since $f(I)$ is connected which is a interval with two endpoint specified hence $J = [y_0,y_1]$. hence any $w_1,w_2 \in I$ ($w_1<w_2$)exist some $k_1,k_2$ maps to it (by surjective of $f$)

Now we assume $k_1>k_2$ we get contradiction which means that $f$ is monotone increasing,hence complete the proof.

To see the contradiction we just need consider the interval $[x_0,w_1]$ which has range $[y_0,k_1]$ hence exist some point in between gets the value $k_2$,now we also has $f(w_2) = k_2$ such that $w_2$ different form this point hence contradiction.

Is my proof correct ,is there some better proof?

yi li
  • 5,414

1 Answers1

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When you assume that $y_{0}<y_{1}$, I would add something like "with this assumption, the problem becomes proving that the function is increasing. In the other hand, if $y_{1}>y_{0}$, the problem is to show that $f$ is decreasing, and the proof is similar". You also have some typos (e.g., when you say that $w_{1},w_{2} \in J$ it should be $w_{1},w_{2} \in I$).

A more direct way is the following: since $f:I \to J$ is a difeomorfism, its derivative $f'$ is never zero (otherwise $(f^{-1})'=1/f'$ will not be defined). Then, $f>0$ in $I$ or $f<0$ in $I$. In the first case, $f$ is increasing while in the second case $f$ is decreasing.

Sebathon
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