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Let $g : X \rightarrow Y$ be a homeomorphism, and let $f : Y \rightarrow \mathbb{R}$ be a continuous function. Then the extremes of $f$ in $Y$ are mapped to the extremes of the composition $f \circ g $ in $X$, and if $f$ has a unique minimizer in $Y$, then $f \circ g $ has a unique minimizer in $X$. This seems plausible to me, but does anyone know of a reference where this is proved (and preferably related results)?

Kind regards


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In fact, you only need $g$ to be a bijection. – Martin Sleziak Jul 29 '12 at 10:39
Yeah that makes sense. Thanks. – Olav Jul 29 '12 at 10:46
up vote 0 down vote accepted

This is one of the things that are so plausible that a formal proof makes them less so. Therefore I shall describe in so many words how I perceive the situation.

From the definition of "homeomorphism" we need only that $g:\ X\to Y$ is bijective. This means that we have a set $$S:=\bigl\{(x,y)\ \bigm|\ x\in X,\ y=g(x)\in Y\bigr\}$$ of "married pairs". Each point $p=(x,y)\in S$ can be adressed either by its first coordinate $x\in X$ or its second coordinate $y\in Y$, and the only difference between $f$ and $f\circ g$ is whether the function $$F:\ S\to {\mathbb R},\quad p=(x,y)\mapsto F(p):=f(y)=f\bigl(g(x)\bigr)$$ is expressed as $F(p):=f(y)$ or as $F(p):=f\bigl( g(x)\bigr)$.

Looking at the setup in this way both your statements become obvious: The extremes of $f$ resp. $f\circ g$ are the extremes of the encompassing idea $F$.

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Thanks for your reply!. Sorry I cant upvote due to lack of points myself. – Olav Jul 29 '12 at 15:18
@Olav But you can accept the answer by clicking the checkmark. – user31373 Jul 31 '12 at 1:51
Yes, that's true. I was waiting a bit to see if someone would come up with a reference. (Is that bad manner?). I've now accepted the answer. – Olav Jul 31 '12 at 12:04
This is way too late, but for the benefit of future readers: No, it's not bad manner to wait for a few days in hopes of multiple answers. – Eric Stucky May 23 at 12:28

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