Understanding Function Composition and Properties

This problem involves understanding the properties of function composition, specifically when dealing with onto (surjective) functions.

Definitions

• Onto (Surjective) Function: A function $h: X \to Y$ is called onto if every element in the codomain $Y$ has at least one corresponding element in the domain $X$. In other words, for every $y \in Y$, there exists at least one $x \in X$ such that $h(x) = y$. The image of the function is equal to its codomain.
• One-to-one (Injective) Function: A function $h: X \to Y$ is called one-to-one if distinct elements in the domain $X$ map to distinct elements in the codomain $Y$. In other words, if $x_1 \neq x_2$, then $h(x_1) \neq h(x_2)$ for all $x_1, x_2 \in X$.
• Bijective Function: A function is bijective if it is both one-to-one (injective) and onto (surjective).

Analyzing the Given Conditions

We are given the following:

• A function $g: A \to B$.
• A function $f: B \to C$.
• The composition function $f \circ g: A \to C$, defined as $f(g(x))$.
• Both $f$ and $f \circ g$ are onto (surjective) functions.

We need to determine the necessary properties of function $g$.

Investigating the Properties of Function $g$

Can $g$ be non-onto?

Let's consider if $g$ *must* be onto. Suppose $g$ is not onto. This means there is at least one element in $B$ that is not the image of any element from $A$. Can $f \circ g$ still be onto? Yes.

Example:

• Let $A = \{1\}$, $B = \{1, 2\}$, $C = \{1\}$.
• Define $g: A \to B$ as $g(1) = 1$. Here, $g$ is not onto because $2 \in B$ has no pre-image in $A$.
• Define $f: B \to C$ as $f(1) = 1$ and $f(2) = 1$. Here, $f$ is onto because for $1 \in C$, we have $f(1)=1$ (and $f(2)=1$).
• Now consider the composition $f \circ g: A \to C$. We have $f(g(1)) = f(1) = 1$. The function $f \circ g$ maps $1$ to $1$. Since the domain $A$ has only one element and the codomain $C$ has only one element, and the mapping $1 \mapsto 1$ covers the codomain, $f \circ g$ is onto.

In this scenario, $f$ is onto, $f \circ g$ is onto, but $g$ is not onto. Therefore, $g$ is not required to be an onto function.

Can $g$ be non-one-to-one?

Let's consider if $g$ *must* be one-to-one. Suppose $g$ is not one-to-one. This means there exist two different elements in $A$ that map to the same element in $B$. Can $f \circ g$ still be onto? Yes.

Example:

• Let $A = \{1, 2\}$, $B = \{1\}$, $C = \{1\}$.
• Define $g: A \to B$ as $g(1) = 1$ and $g(2) = 1$. Here, $g$ is not one-to-one because $g(1) = g(2)$ but $1 \neq 2$.
• Define $f: B \to C$ as $f(1) = 1$. Here, $f$ is onto.
• Now consider the composition $f \circ g: A \to C$. We have $f(g(1)) = f(1) = 1$ and $f(g(2)) = f(1) = 1$. The function $f \circ g$ maps both $1$ and $2$ to $1$. Since the only element in the codomain $C$ is $1$, and both elements of $A$ map to it, $f \circ g$ is onto.

In this scenario, $f$ is onto, $f \circ g$ is onto, but $g$ is not one-to-one. Therefore, $g$ is not required to be a one-to-one function.

Conclusion

Since we have shown examples where $g$ can be neither onto nor one-to-one while still satisfying the conditions that $f$ and $f \circ g$ are onto, the function $g$ is not required to be one-to-one or onto.

The correct statement is that $g(.)$ is not required to be a one-to-one or onto function.