Munkres - Functions

Adar Kahiri
June 2020

Problem 1: Let \(f: A \rightarrow B\). Let \(A_0 \subset A\) and \(B_0 \subset B\).


Let's start off by defining \(f\) and \(f^{-1}\). Given \(f: A \rightarrow B\), \(A_0 \subset A\), and \(B_0 \subset B\), \[f(A_0) = \{b \ | \ b=f(a) \ \textrm{for at least one}\ a \in A_0\}\] \[f^{-1}(B_0)=\{ a \ | \ f(a) \in B_0 \}\] Now, to answer part (a), if the function \(f\) is not injective, then it's possible that there are two elements \(a\) and \(a'\) such that \[f(a)=f(a') \ \textrm{and} \ a \in A_0 \ \textrm{and} \ a' \not\in A_0\] And thus, the preimage of \(f(A_0)\), \(f^{-1}(f(A_0))\), could contain elements not in \(A_0\).

However, if \(f\) is injective, then \[[f(a) = f(a')] \implies [a=a']\] and so, any element in the preimage would have to be an element of \(A_0\), meaning the equality would hold.

Similarly, if \(f\) is not surjective, then it's possible that there exists an element \(b \in B\) to which there isn't a mapping from A, and thus \[b \in f(f^{-1}(B_0)) \implies b \in B\] but \[b \in B \not\Longrightarrow b \in f(f^{-1}(B_0))\] meaning that \(f(f^{-1}(B_0)) \subset B_0\).

However, if \(f\) is surjective, then there must be a mapping to every element of \(B\) from \(A\), meaning that \[[b \in f(f^{-1}(B_0)) \iff b \in B] \iff [f(f^{-1}(B_0)) = B]\]