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Let $X$ and $Y$ be topological spaces, and $h:X \to Y$ a homeomorphism. For every continuous map $f:X \to X$, is there a continuous map $g:Y \to Y$ such that $f=h^{-1} \circ g \circ h$?

This came up in trying to prove that if $X$ is homeomorphic to $D^n$, the $n$-dimensional disk, then every continuous function $f:X \to X$ has a fixed point. If the above question can be answered in the affirmative, then there is an easy proof using that plus Brouwer's fixed point theorem.

Thanks.

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    $\begingroup$ Doesn't taking $g = h \circ f \circ h^{-1}$ immediately give you the result? $\endgroup$ – André 3000 Mar 25 '14 at 3:38
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As SpamIAm said, the map $g=h\circ f\circ h^{-1}$ does the job. The intuition here is that if $X$ and $Y$ are spaces which are homeomorphic and $Z$ is some other space, then a homeomorphism from $X$ to $Y$ induces a bijection $\operatorname{Hom}(X,Z)\leftrightarrow\operatorname{Hom}(Y,Z)$ (here $\operatorname{Hom}(X,Z)$ is the set of continuous maps rom $X$ to $Z$) and a bijection $\operatorname{Hom}(Z,X)\leftrightarrow\operatorname{Hom}(Z,Y)$. In other words, homeomorphic spaces $X$ and $Y$ are the same in the categorical sense: the sets of maps out of each space (and into each space) are more or less the same.

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