GATE CS 2025 Set 1 Q7 — Sets, Relations and Functions

To determine the properties of the function $g: A \to B$, we can construct a simple counterexample where both $f$ and $f \circ g$ are onto, but $g$ is neither one-to-one (injective) nor onto (surjective).

Let the sets be:

• $A = \{1, 2, 3\}$
• $B = \{x, y, z\}$
• $C = \{p, q\}$

Define the function $g: A \to B$ as:

• $g(1) = x$
• $g(2) = x$
• $g(3) = y$

Here, $g$ is not one-to-one (since $g(1) = g(2) = x$) and not onto (since $z \in B$ has no preimage in $A$).

Define the function $f: B \to C$ as:

• $f(x) = p$
• $f(y) = q$
• $f(z) = q$

Here, $f$ is onto because the range of $f$ is $\{p, q\} = C$.

Now, let us find the composition $(f \circ g): A \to C$:

• $(f \circ g)(1) = f(g(1)) = f(x) = p$
• $(f \circ g)(2) = f(g(2)) = f(x) = p$
• $(f \circ g)(3) = f(g(3)) = f(y) = q$

The composition $f \circ g$ is onto because its range is $\{p, q\} = C$.

Since both $f$ and $f \circ g$ are onto, but $g$ is neither one-to-one nor onto, we conclude that $g$ is not required to be one-to-one or onto.