# If $X$ is a topological space homeomorphic to $D^n$ then every continuous $f:X\to X$ has a fixed point. [closed]

This is Rotman "Intro to Algebraic Topology" Exercise 0.4:

If $$X$$ is a topological space homeomorphic to $$D^n$$, then every continuous $$f:X\to X$$ has a fixed point.

How can this be shown?

• Did Rotman prove that it is true for $X = D^n$? Commented May 2, 2023 at 22:23
• @PaulFrost Yes, it is given as Theorem 0.3 of his book, attributed to Brouwer. Commented May 2, 2023 at 22:36

Let $$f:X\to X$$ be a continuous function. Since $$X \cong D^n$$, we have a homeomorphism (i.e., a continuous bijection with continuous inverse) $$h:X \to D^n$$. Consider the diagram $$D^n \xrightarrow{h^{-1}} X \xrightarrow{f} X \xrightarrow h D^n,$$ and observe that the composition $$g := h \circ f \circ h^{-1}$$ is continuous as the composition of continuous functions. Therefore by Brouwer's Fixed Point Theorem, there exists $$x\in D^n$$ such that $$g(x)=x$$. As a consequence, $$f(h^{-1}(x))= h^{-1}(x)$$, so $$f$$ also has a fixed point.