(May 2016) Suppose that the space \(X\) has the fixed point property (that is, for any continuous function \(f\colon X\to X\) there is a point \(p\in X\) with \(f(p)=p\)). Suppose also that \(A\subset X\) is a subspace admitting a retraction \(r\colon X\to A\). Show that \(A\) also has the fixed point property.
Solution
Let \(f:A\to A\) be continuous, and let \(i:A\hookrightarrow X\) be inclusion. Define \[ F=i\circ f\circ r:X\longrightarrow X. \] Since \(X\) has the fixed point property, there is \(x\in X\) with \(F(x)=x\). But \(F(x)\) lies in \(i(A)=A\), so \(x\in A\). Since \(r\) is a retraction, \(r(x)=x\). Therefore \[ x=F(x)=i(f(r(x)))=f(x), \] where we identify \(i(A)\) with \(A\). Thus \(f\) has a fixed point. Since \(f\) was arbitrary, \(A\) has the fixed point property.