[Math] If $fg$ is surjective, then $g$ is surjective.

functionsreal-analysis

Either prove or give a counterexample to the "converses" of exercise 2 on page 17

  1. If $fg$ is injective, $f$ is injective.
  2. If $fg$ is injective, $g$ is injective.
  3. If $fg$ is surjective, $f$ is surjective.
  4. 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?

Best Answer

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 $g(x) = 2$.

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