I know that this property is basic, but I can't demonstrate it.
If $A$ is a retract of an space X that has the fixed point property (f.p.p), then, $A$ has the f.p.p, so what I have to show is that for all $g\in\mathcal{C}(A,A)$ exists $a\in A$ that $g(a)=a$.
I know that for all $f\in\mathcal{C}(X,X)$ exists $x\in X$ that $f(x)=x$. As $A$ is a retract, we have that $r\circ i=id_{A}$, $r\in\mathcal{C}(X,A)$, so my problem is, what $a\in A$ I have to take? My intuition is to take the $x\in X$ for which I have the f.p.p for $X$ and apply to it $r$.
Thanks for the answers!
Best Answer
Let $r:X\to A$ be a retraction and $i:A\to X$ the inclusion. Assume that $f:A\to A$ is a continuous map and consider $i\circ f\circ r$ which is a continuous map $X\to X$. Since $X$ has FPP then there is $a\in X$ such that $i\circ f\circ r(a)=a$. But the image $\text{im}(i\circ f\circ r)$ is a subset of $A$ (because $\text{im}(i)=A$) and thus $a\in A$. Now $i(f(x))=f(x)$ obviously and since $r$ is a retraction then $r(a)=a$. It follows that $f(a)=a$ and so $a$ is a fixed point of $f$.