The Schauder fixed point theorem states that if $X$ is a Banach space, $K\subset X$ is a convex, bounded and closed subset and $T:K\rightarrow K$ is compact, then $T$ has, at least, one fixed point in $K$. I want to know if this statement still holds true when the subset $K$ is replaced by another subset $D$ that is homeomorphism to $K$.
Roughly speaking, if $h:K\rightarrow D$ is an homeomorphism, then $h^{-1}Th:K\rightarrow K$ and I am able to apply the Schauder fixed point theorem if $h^{-1}Th$ is compact.
Is there any result that guarantee that the compactness of operators is preserved by homeomorphism or something similar?
If $T(D)$ is a compact subset of $D$, then $Th(K)=T(D)$ is a compact subset of $D$, whence $h^{-1}Th(D)$ is a compact subset of $K$. Thus there is some $x_0\in K$ such that $h^{-1}Th(x_0)=x_0$. Put $y_0=h(x_0)$. Then $Ty_0=hh^{-1}Th(x_0)=h(x_0)=y_0$, so $T$ has a fixed point.