Piecewise Composite Functions and Surjective/Injective I understand how to work out a composite function when it is just one function but have no clue when it comes to piecewise composite functions like follows:
$$  f(x) =
\begin{cases}
x-1,  & \ x\ge0 \\
x^3, & \ x < 0
\end{cases}
$$
$$  g(x) =
\begin{cases}
x+1,  & \ x\ge0 \\
2x, & \ x < 0
\end{cases}
$$
How would I go about working out  $ f \circ g $ and $g \circ f$?
Also, how can I tell if a piece wise function is surjective or injective? By graphing both of them I have worked out (if I am correct) that $f(x)$ is not injective as $f(0) = f(-1)$ and $g(x)$ is injective. 
Thanks in advance.
 A: You are correct that $f$ is not injective and that $g$ is injective.  
The graph of $f$ is shown below.
 
The function assumes all negative values in the interval $(-\infty, 0)$ and all values greater than or equal to $-1$ in the interval $[0, \infty)$.  Consequently, it assumes every real value, so it is surjective.
The graph of $g$ is shown below.

Since the function $g$ assumes all values greater than or equal to $1$ in the interval $[0, \infty)$ and all negative values in the interval $(-\infty, 0)$, it never assumes the values in the interval $[0, 1)$, so $g$ is not surjective.
Let's find $f \circ g$.  By definition, $(f \circ g)(x) = f(g(x))$.  Since the rule for $f$ is determined by whether the value substituted into the function is nonnegative or negative, we have to determine whether $g(x)$ is nonnegative or negative in order to determine which rule to apply.  
If $x \geq 0$, $g(x) = x + 1 \geq 1$, so we are applying $f$ to positive values.  Hence, we apply the rule $f(x) = x - 1$ to $g(x) = x + 1$.  Thus, if $x \geq 0$, 
\begin{align*}
(f \circ g)(x) & = f(g(x))\\
               & = f(x + 1)\\
               & = x + 1 - 1\\
               & = x  
\end{align*}
If $x < 0$, then $g(x) = 2x < 0$, so we are applying $f$ to negative values.
Hence, we apply the rule $f(x) = x^3$ to $g(x) = 2x$.  Thus, if $x < 0$,
\begin{align*}
(f \circ g)(x) & = f(g(x))\\
               & = f(2x)\\
               & = (2x)^3\\
               & = 8x^3
\end{align*}
Hence, 
$$(f \circ g)(x) = 
\begin{cases}
x & \text{if $x \geq 0$}\\
8x^3 & \text{if $x < 0$}
\end{cases}
$$
The graph of the composite function $f \circ g$ is shown below.
 
From the graph, we see that $f \circ g$ assumes every real value precisely once.  Hence, $f \circ g$ is both injective and surjective.  
I will leave finding $g \circ f$ to you.  Be careful.  Notice that $f$ changes sign at $x = 1$.  Thus, you will need to consider three intervals, $[1, \infty)$, $[0, 1)$, and $(-\infty, 0)$.  As @EthanBolker suggested in the comments, drawing the graph of the composite function is helpful in determining whether it is injective or surjective.
