If $fg$ is surjective, then $g$ is surjective. Either prove or give a counterexample to the "converses" of exercise 2 on page 17


*

*If $fg$ is injective, $f$ is injective.

*If $fg$ is injective, $g$ is injective.

*If $fg$ is surjective, $f$ is surjective.

*If $fg$ is surjective, $g$ is surjective.


$\textbf{Part 1 (Counterexample):}$ Let $g(x)=1$ and $f(x)=x$ for all x's.
So assume fg is injective. (i.e. $fg(x_1)=fg(x_2) \rightarrow x_1=x_2$)
\begin{equation*}
\begin{aligned}
fg(x_1)=fg(x_2) & \rightarrow f(g(x_1))=f(g(x_2)) \\
& \rightarrow f(x_1)=f(x_2)\\
& \rightarrow 1=1 \\
\end{aligned}
\end{equation*}
However because $f(x)=1$ we can have two different x's but still return the same answer, 1. Hence f is not injective.
$\textbf{Part 2:(Counterexample):}$ Let $g(x)=1$ and $f(x)=x$ for all x's.
So assume fg is injective. (i.e. $fg(x_1)=fg(x_2) \rightarrow x_1=x_2$)
\begin{equation*}
\begin{aligned}
fg(x_1)=fg(x_2) & \rightarrow f(g(x_1))=f(g(x_2)) \\
& \rightarrow f(x_1)=f(x_2)\\
& \rightarrow 1=1 \\
\end{aligned}
\end{equation*}
However because $g(x)=1$ we can have two different x's but still return the same answer, 1. Hence g is not injective.
$\textbf{Part 3:}$ Let $f:A \to B$ and $g:B \to C$. Assume fg is surjective. Since $f(g(x))$ is surjective, for all $a \in A$ there is a $c \in C$ such that $f(g(c))=a$. But since $g(c) \in C$ (by definition of g), that means for all $a \in A$, there is a $b \in B$ (namely g(c) such that f(b)=a). So f is surjective.
$\textbf{Part 4:}$ How would I amend the proof for Part 3 for Part 4?
 A: Q1. False. But your counterexample is invalid because your $fg$ is not injective. E.g. $f \circ  g(0) = f(1) = 1$ and $f \circ g(1) = f(1) = 1$.
Instead, define the functions:
$$g:[0,1] \rightarrow [0,2] \mbox{ by } g(x) = x$$
$$f:[0,2] \rightarrow [0,1] \mbox{ by } f(x) =
\begin{cases}
x  & \text{if } 0 \leq x \leq 1 \\
x-1 & \text{if } 1 \lt x \leq 2\end{cases}
$$
For function $fg:[0,1] \rightarrow [0,1],\,$ we have $ f\circ g(x) = x,\,\, \forall\, x \in [0,1]$ so it is clearly injective but $f$ is not injective because, for example, $f(2) = 1 = f(1)$.
$\\$
Q2. True. We prove it by contradiction. Assume $fg$ is injective and suppose $\exists\,\, x,y \in Dom(g),\,\, x \neq y$, such that $g(x) = g(y)$ so that $g$ is not injective.
Now, $g(x) = g(y)$ implies $f \circ g(x) = f \circ g(y)$ but then $x \neq y$ contradicts $fg$ being injective. Thus, $g$ must be injective.
$\\$
Q3. True. Let $x \in Cod (f)$. Since $fg$ is surjective, $\exists\,\, y \in Dom (g)$ such that $f(g(y)) = x$. But $g(y) \in Dom (f)$ so $f$ is surjective.
$\\$
Q4. False. We use the same functions in $Q1$ as a counterexample.
$fg:[0,1] \rightarrow [0,1]$ is surjective: if $x \in Cod(f) = [0,1]$, then $f\circ g(x) = x$.
$g:[0,1] \rightarrow [0,2]$ is not surjective since $\not\exists\,\, x \in [0,1]$ such that $g(x) = 2$.
