Determining Injectivity, surjectivity, bijectivity, and inverses I was given a question that begins like this. 
Suppose that $A$ is the set $\{a,b,c\}$ (these are just names for some three elements - you don't know anything about $a,b,$ or $c$).  Consider the following functions:


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*(I) Let $f:A \to A$ be the function defined by: $f(a) = c, f(b) = c, f(c) = c$. And let $g:A \to A$ be the function defined by: $g(a) = a, g(b) = b, g(c) = a$.

*(II) Let $f:A \to A$ be the function defined by: $f(a) = b, f(b) = b, f(c) = a$. And let $g:A \to A$ be the function defined by: $g(a) = c, g(b) = a, g(c) = b$.

*For each of these pairs:
(a) Determine for each function whether it is injective, surjective, both or neither.
(b) Determine formulas for $f\circ f$, $g \circ g $, $f \circ g$ and $g \circ f$ for each pair.
(c) Determine which of the functions in part (b) are injective, surjective, both or neither.
(d) For any of the functions from (a) or (b) which are bijections, determine the inverse of that function (i.e. produce a formula for it).
Here is my observation
A) For function $f: A \rightarrow A$  It is not injective or surjective because it is not one to one or onto since the function maps to element c. For the function $g: A \rightarrow A$ is surjective because every element y in Y has a corresponding element x in X such that f(x) = y.
B) I did not understand what $f\circ f$, $g \circ g $, $f \circ g$ and $g \circ f$ mean, this is what I need clarification on.
C) I thought $f: A \rightarrow A$ was surjective not injective, and $g: A \rightarrow A$ is bijective.
D)For D I was confused on how to determine an inverse because I'm use to functions such as f(x) = 2x etc.
 A: On your question B: If one has a function $f: A \to A$ and one $g: A \to A$ then one can compose these functions because the domain of one contains the image of the other. For instance if you have $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$ and $g: \mathbb{R} \to \mathbb{R}$ given by $g(x) = (x-1)$ you can define the composition $f \circ g: \mathbb{R} \to \mathbb{R}$ given by $(f \circ g)(x) = (x-1)^2$ (first we feed the variable $x$ to $g$ which gives us $(x-1)$, and then we feed that to $f$ which squares it). So for instance in your first problem $(f \circ f)(x) = c$ for all $x \in A$ because $f(x) = c$ and $(f \circ f)(x) = f(f(x)) = f(c) = c$.
On questions A, C: Your answers are correct, but the following general principle is worth noting. If you have a function $f: A \to B$ where $A, B$ are finite sets of the same size, then $f$ is bijective if and only if it is injective if and only if it is surjective. This is a simple counting argument that might be enlightening for you to try to carry out yourself.
On your question D: There will not be an easy polynomial formula for $f^{-1}, g^{-1}$ because multiplication and addition are not defined on $A$, you simply have to write out what its value is on each element of $A$ as in the problem statement.
