Section 9: Problem 5 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

(a) Use the choice axiom to show that if $f:A\rightarrow B$ is surjective, then $f$ has a right inverse $h:B\rightarrow A$ .

(b) Show that if $f:A\rightarrow B$ is injective and $A$ is not empty, then $f$ has a left inverse. Is the axiom of choice needed?

(a) If $B$ is empty, then the statement is vacuously true, otherwise $A$ is not empty ( $f$ is surjective). Consider the collection $\mathcal{A}$ of nonempty sets $\{f^{-1}(y)\}_{y\in B}$ ( $f$ is surjective), and let $c$ be a choice function for $\mathcal{A}$ . For every $y\in B$ , let $h(y)=c(f^{-1}(y))$ . Then $(f\circ h)(y)=y$ for every $y\in B$ .

(b) The axiom of choice is not needed. Let $a_{0}\in A$ . For $b\in f(A)$ , there is unique $a\in A$ such that $b=f(a)$ , so let $g(b)=a$ . For $b\notin f(A)$ , let $g(b)=a_{0}$ . We have $(g\circ f)(a)=a$ for all $a\in A$ .

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Section 9: Problem 5 Solution

Section 9: Problem 5 Solution