## Functions

June 2020

**Problem 1: 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.**

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

\[[b \in f(f^{-1}(B_0)) \iff b \in B] \iff [f(f^{-1}(B_0)) = B]\]

\[\tag*{$\blacksquare$}\]