$g(.)$ is a function from $A$ to $B$, $f(.)$ is a function from $B$ to $C$, and their composition defined as $f(g(.))$ is a mapping from $A$ to $C$. If $f(.)$ and $f(g(.))$ are onto (surjective) functions, which ONE of the following is TRUE about the function $g(.)$?

If the composite function $f \circ g: A \to C$ is onto (surjective), it guarantees that the outer function $f: B \to C$ is onto. However, it imposes no requirement on the inner function $g: A \to B$ to be injective (one-to-one) or surjective (onto).

### Counterexample: Let the sets be defined as:

• $A = \{1, 2\}$

• $B = \{x, y, z\}$

• $C = \{c\}$

Define the mappings:

• $g(1) = x$, $g(2) = x$

• $f(x) = c$, $f(y) = c$, $f(z) = c$

Checking the properties:

1. $f$ is onto: $\text{range}(f) = \{c\} = C$.

2. $f \circ g$ is onto: $(f \circ g)(1) = f(x) = c$ and $(f \circ g)(2) = f(x) = c$, so $\text{range}(f \circ g) = \{c\} = C$.

3. $g$ is NOT one-to-one: $g(1) = g(2) = x$ with $1 \neq 2$.

4. $g$ is NOT onto: The elements $y, z \in B$ are not mapped from any element in $A$.

Thus, $g(.)$ is not required to be either one-to-one or onto.

Therefore, (D) is the correct statement.