GATE CS 2005 Q43 — Sets, Relations and Functions

We are given that $h = f \circ g : A \to C$ is an onto (surjective) function. This means that for every element $c \in C$, there exists some $a \in A$ such that:

Since $g(a) \in B$, let $b = g(a)$. Then $f(b) = c$. This implies that for every $c \in C$, there exists an element $b \in B$ such that $f(b) = c$. Therefore, $f$ must be onto.

However, $g$ does not need to be onto. Consider the following counterexample:

• Let $A = \{1\}$, $B = \{x, y\}$, and $C = \{z\}$.
• Define $g: A \to B$ as $g(1) = x$ (not onto, since $y$ is not mapped).
• Define $f: B \to C$ as $f(x) = z$ and $f(y) = z$ (onto).
• Then, $h(1) = f(g(1)) = f(x) = z$, which makes $h: A \to C$ onto.

Thus, $f$ must be onto, but $g$ need not be onto.