Section 2: Problem 2 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

Let $f:A\rightarrow B$ and let $A_{i}\subset A$ and $B_{i}\subset B$ for $i=0$ and $i=1$ . Show that $f^{-1}$ preserves inclusions, unions, intersections, and differences of sets:

(a) $B_{0}\subset B_{1}\Rightarrow f^{-1}(B_{0})\subset f^{-1}(B_{1})$ .

(b) $f^{-1}(B_{0}\cup B_{1})=f^{-1}(B_{0})\cup f^{-1}(B_{1})$ .

(d) $f^{-1}(B_{0}-B_{1})=f^{-1}(B_{0})-f^{-1}(B_{1})$ .

Show that $f$ preserves inclusions and unions only:

(e) $A_{0}\subset A_{1}\Rightarrow f(A_{0})\subset f(A_{1})$ .

(f) $f(A_{0}\cup A_{1})=f(A_{0})\cup f(A_{1})$ .

(g) $f(A_{0}\cap A_{1})\subset f(A_{0})\cap f(A_{1})$ ; show that equality holds if $f$ is injective.

(h) $f(A_{0}-A_{1})\supset f(A_{0})-f(A_{1})$ ; show that equality holds if $f$ is injective.

(a) $x\in f^{-1}(B_{0})\Rightarrow$ $f(x)\in B_{0}\Rightarrow$ $f(x)\in B_{1}\Rightarrow$ $x\in f^{-1}(B_{1})$ .

(b) Using (a), $f^{-1}(B_{i})\subset f^{-1}(B_{0}\cup B_{1})$ , and, therefore, $f^{-1}(B_{0})\cup f^{-1}(B_{1})\subset f^{-1}(B_{0}\cup B_{1})$ . The other direction: $x\in f^{-1}(B_{0}\cup B_{1})\Rightarrow$ $f(x)\in B_{0}$ or $f(x)\in B_{1}\Rightarrow$ $x\in f^{-1}(B_{0})\cup f^{-1}(B_{1})$ .

(c) Using (a), $f^{-1}(B_{0}\cap B_{1})\subset f^{-1}(B_{i})$ , and, therefore, $f^{-1}(B_{0}\cap B_{1})\subset f^{-1}(B_{0})\cap f^{-1}(B_{1})$ . The other direction: $x\in f^{-1}(B_{0})\cap f^{-1}(B_{1})\Rightarrow$ $f(x)\in B_{0}$ and $f(x)\in B_{1}\Rightarrow$ $x\in f^{-1}(B_{0}\cap B_{1})$ .

(d) $x\in f^{-1}(B_{0}-B_{1})$ iff $f(x)\in B_{0}-B_{1}$ iff $f(x)\in B_{0}$ and $f(x)\notin B_{1}$ iff $x\in f^{-1}(B_{0})$ and $x\notin f^{-1}(B_{1})$ iff $x\in f^{-1}(B_{0})-f^{-1}(B_{1})$ .

(e) $y\in f(A_{0})\Rightarrow$ $\exists x\in A_{0}\subset A_{1}$ : $f(x)=y\Rightarrow$ $y\in f(A_{1})$ .

(f) Using (e), $f(A_{i})\subset f(A_{0}\cup A_{1})\Rightarrow$ $f(A_{0})\cup f(A_{1})\subset f(A_{0}\cup A_{1})$ . The other direction: $y\in f(A_{0}\cup A_{1})\Rightarrow$ $\exists x\in A_{0}\cup A_{1}$ : $f(x)=y\Rightarrow$ $\exists x\in A_{0}$ : $f(x)=y$ or $\exists x\in A_{1}$ : $f(x)=y\Rightarrow$ $y\in f(A_{0})\cup f(A_{1})$ .

(g) The " $\subset$ " part follows from (e). If $f$ is injective then $y\in f(A_{0})\cap f(A_{1})\Rightarrow$ $\exists x\in A_{0}$ : $f(x)=y$ and $\exists x'\in A_{1}$ : $f(x')=y$ . Since $f$ is injective, $x=x'\in A_{0}\cap A_{1}\Rightarrow$ $y\in f(A_{0}\cap A_{1})$ .

(h) If $y\in f(A_{0})-f(A_{1})$ , then $\exists x\in A_{0}$ : $f(x)=y$ and, hence, $x\notin A_{1}$ . Therefore, $y\in f(A_{0}-A_{1})$ . If $f$ is injective then $y\in f(A_{0}-A_{1})\Rightarrow$ $\exists x\in A_{0}-A_{1}$ : $f(x)=y\Rightarrow$ $y\in f(A_{0})$ and $y\notin f(A_{1})$ (otherwise, $y=f(x')$ for some $x'\in A_{1}$ , and, since $f$ is injective, $x=x'\in A_{1}$ ) $\Rightarrow$ $y\in f(A_{0})-f(A_{1})$ .

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Section 2: Problem 2 Solution

Section 2: Problem 2 Solution