## 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

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

$\tag*{\blacksquare}$