Naive Set Theory: Functions
Function Classes
Functions can be classified based on some properties of how often elements in their domains map to elements in their ranges.
A function $f : A \rightarrow B$ is injective if $f(a) = f(b)$ implies that $a = b$ for any two elements $a$ and $b$ in $A$. An injective function is also said to be onetoone, since each element in the codomain is mapped to by at most one element in the domain. An injective function is called an injection. Pretty intense, right? 

A function $f : A \rightarrow B$ is surjective if for each element $b \in B$ there exists at least one $a \in A$ such that $f(a) = b$. In other words, a function is surjective if all the elements in its codomain are mapped to by at least one element in its domain. A surjective function is also said to be onto, although this term is harder to remember. Most people think of it as "the other special one that's not onetoone." A surjective function is also called a surjection. 

A function is bijective if it is both injective and surjective. A bijective function is also called a (wait for it) bijection. 

There is no special name for a function that is neither an injection nor a surjection. Perhaps we could call it a nonjection? 
Problems
Let $f : A \rightarrow B$ be an injective function, and let $A' \subseteq A$. Prove that $A' = f^{1}(f(A'))$.
First we show that $A' \subseteq f^{1}(f(A'))$. This was shown in the first problem in the Images and Preimages section.
Next we show that $f^{1}(f(A')) \subseteq A'$. Let $a \in f^{1}(f(A'))$. Because $f$ is injective, there is a unique $f(a) \in f(A')$. As a result $a \in A$. Therefore, $f^{1}(f(A')) \subseteq A'$.
Let $f : A \rightarrow B$ be a surjective function, and let $B' \subseteq B$. Prove that $B' = f(f^{1}(B))$.
First we show that $B' \subseteq f(f^{1}(B))$. Let $b \in B'$. Because $f$ is surjective, there is at least one $a \in A$ such that $b = f(a)$. Then $a \in f^{1}(B)$. It follows that $f(a) \in f(f^{1})(B)$, so $b \in f(f^{1}(B))$. Therefore $B' \subseteq f(f^{1}(B))$
The second part of this proof showing $f(f^{1}(B)) \subseteq B'$ is given in the second problem in the Images and Preimages section.