Section 2: 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

In general, let us denote the identity function for a set $C$ by $i_{C}$ . That is, define $i_{C}:C\rightarrow C$ to be the function given by the rule $i_{C}(x)=x$ for all $x\in C$ . Given $f:A\rightarrow B$ , we say that a function $g:B\rightarrow A$ is a left inverse for $f$ if $g\circ f=i_{A}$ ; and we say that $h:B\rightarrow A$ is a right inverse for $f$ if $f\circ h=i_{B}$ .

(a) Show that if $f$ has a left inverse, $f$ is injective; and if $f$ has a right inverse, $f$ is surjective.

(b) Give an example of a function that has a left inverse but no right inverse.

(d) Can a function have more than one left inverse? More than one right inverse?

(e) Show that if $f$ has both a left inverse $g$ and a right inverse $h$ , then $f$ is bijective and $g=h=f^{-1}$ .

(a) Apply 4 (c) and (e) using the fact that the identity function is bijective.

(b) $e^{x}:\mathbb{R}\rightarrow\mathbb{R}-\{0\}$ has at least two left inverses $ln(|x|)$ and, for example, $ln(|max\{x,-1\}|)$ but no right inverses (it is not surjective).

(c) $x^{2}:\mathbb{R}\rightarrow\overline{\mathbb{R}}_{+}$ has two right inverses $\sqrt{x}$ and $-\sqrt{x}$ but no left inverse (it is not injective).

(d) See (b) and (c).

(e) It follows from (a) that $f$ is bijective, i.e. for each $y\in B$ there is unique $x\in A$ such that $f(x)=y$ . Now, $g(y)=g(f(x))=x$ , and $f(x)=y=f(h(y))$ , and, since $f$ is injective, $h(y)=x$ as well.

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Section 2: Problem 5 Solution

Section 2: Problem 5 Solution