Question

Consider function f: A→B and g: B→C (A, B, C⊆R) such that (gof)⁻¹ exists then:

• A. f and g both are one-one
• B. f and g both are onto
• C. f is one-one and g is onto
• D. f is onto and g is one-one

Answer 1

Answer: (C)

f is one-one and g is onto

Answer 2

The existence of $(g \circ f)^{-1}$ implies that the composite function $g \circ f$ is bijective (both one-one and onto).

1. Injectivity of $g \circ f$: If $g \circ f$ is injective, then $f$ must be injective. This is because if $f(a_1) = f(a_2)$, then $(g \circ f)(a_1) = g(f(a_1)) = g(f(a_2)) = (g \circ f)(a_2)$. For $g \circ f$ to be injective, $a_1 = a_2$ must hold, which means $f$ must be one-one.

2. Surjectivity of $g \circ f$: If $g \circ f$ is surjective, then for every $c \in C$, there exists an $a \in A$ such that $(g \circ f)(a) = c$. This means $g(f(a)) = c$. Let $y = f(a)$. Since $f: A \to B$, $y \in B$. Thus, for every $c \in C$, there exists an element $y$ in the domain of $g$ (which is $B$) such that $g(y) = c$. This shows that $g$ is surjective (onto).

Therefore, if $(g \circ f)^{-1}$ exists, then $f$ is one-one and $g$ is onto.

The conditions for the existence of $(g \circ f)^{-1}$ are that $f$ must be injective and $g$ must be surjective.