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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

Olav

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    Yeah that makes sense. Thanks.2012-07-29

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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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    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.2016-05-23