Consider functions $f: A \rightarrow B$ and $g: B \rightarrow C(A, B, C \subseteq R)$ such that $(g \circ f)^{-1}$ exists, then :

(C) $f$ is one-one and $g$ is onto

(B) $f$ is onto and $g$ is one-one

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Sets Relations and Functions

Consider functions

f : A \to B

and

g : B \to C (A, B, C \subseteq R)

such that

(g \circ f)^{- 1}

exists, then :

Answer: (C)

An inverse of

g \circ f : A \to C

exists only if

g \circ f

is both one-one and onto. If

f (x_{1}) = f (x_{2})

for some

x_{1} \neq = x_{2}

, then

(g \circ f) (x_{1}) = (g \circ f) (x_{2})

, so injectivity of the composite forces

f

to be one-one. Every element of

C

equals

g (f (x))

for some

x \in A

, and

f (x) \in B

, so every element of

C

has a pre-image in

B

, which makes

g

onto. So

f

is one-one and

g

is onto.

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option