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.
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.