If $A$ is a retract of $B^2$, then every map $f: A \rightarrow A$ has a fixed point

Question

Let $r$ be the retract. Then $r: B^2 \rightarrow A$ is continuous and $r|_A=Id_A$.

Let $f$ be a map from $A$ to $A$ and let $j$ be in inclusion mapping, i.e $j: A \rightarrow B^2$ where $A \cong j(A)$.

Let $\gamma = j \circ f \circ r$, then $\gamma$ is a map from $B^2$ to $B^2$ and so by Brouwers fixed point theoerem $\exists x\in B^2$ s.t. $\gamma(x)=x$.

Also, note that $\gamma(B^2) \subset A$, so this $x$ that we have found is actually an element of $A$, and furthermore $\gamma|_A(B^2) \cong f(A)$, and so $f(x)=x$.

See what i'm going for? Is there a more standard way to show this? Thanks!

Answer 1

Your proof is fine. I don't know if there is a standard way, but here is a slightly more concise argument: if $A$ is a retract of $D^2$, then there exists $r : D^2 \to A$ with $rj = 1_A$. As you say, $jfr$ has a fixed point $x \in D^2$. Now, since $$ rjfr = (rj)fr = 1_Afr = fr,$$ then $$ f(r(x)) = fr(x) = rjfr(x) = r(jfr(x)) = r(x). $$

and so $f$ fixes $r(x)$.