Section 2: Problem 1 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 $f:A\rightarrow B$ . Let $A_{0}\subset A$ and $B_{0}\subset B$ .

(a) Show that $A_{0}\subset f^{-1}(f(A_{0}))$ and that equality holds if $f$ is injective.

(b) Show that $f(f^{-1}(B_{0}))\subset B_{0}$ and that equality holds if $f$ is surjective.

To show that for some sets $X$ and $Y$ , $X\subset Y$ , we need to show that $\forall x\in X$ , $x\in Y$ , or, equivalently, show that $\forall x\notin Y$ , $x\notin X$ .

(a) $a\in A_{0}\Rightarrow$ $f(a)\in f(A_{0})\Rightarrow$ $a\in f^{-1}(f(A_{0}))$ . If $f$ is injective, then $a\notin A_{0}\Rightarrow$ $f(a)\notin f(A_{0})$ (otherwise, there exists $b\in A_{0}$ such that $b\neq a$ and $f(a)=f(b)$ ) $\Rightarrow$ $a\notin f^{-1}(f(A_{0}))$ .

(b) $y\notin B_{0}\Rightarrow$ $f^{-1}(y)\cap f^{-1}(B_{0})=\emptyset\Rightarrow$ $y\notin f(f^{-1}(B_{0}))$ . If $f$ is surjective, then $f^{-1}(y)\neq\emptyset$ , and $y\in B_{0}\Rightarrow$ $\exists x\in f^{-1}(y)\subset f^{-1}(B_{0})\Rightarrow$ $y\in f(f^{-1}(B_{0}))$ .

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Section 2: Problem 1 Solution

Section 2: Problem 1 Solution