Conditions on $f$ and $g$ if $(g\circ f)^{-1}$ exists Consider functions $f:A\to B$ and $g:B\to C$ $(A,B,C\subseteq R)$ such that $(g\circ f)^{-1}$ exists, then :
(1) $f$ is onto and $g$ is one-one
(2) $f$ is one-one and $g$ is onto
(3) $f$ and $g$ are both one-one
(4) $f$ and $g$ are both onto
One of the above options is supposed to be correct but I think for inverse to exist we must have bijection so both (3) and (4) should be correct. I am little confused
 A: Think of the following simple set functions:
$$\{1,2\} \to \{1,2,3\} \to \{1,2\},$$
where the first one sends $1$ to $1$ and $2$ to $2$, and the second one sends $1$ to $1$, $2$ to $2$ and $3$ to $1$.
The composition is the identity and thus clearly invertible. The functions, though, are not invertible.
You should be able to derive the answer you are looking for from this example.
A: It is easy to prove that
$g\circ f\;$ is one-one $\implies f\;$ is one-one.
$g\circ f\;$ is onto $\implies g\;$ is onto.
Consequently,
since $\;\left(g\circ f\right)^{-1}$ exists, then $\;g\circ f\;$ is one-one and onto, hence $\;f\;$ is one-one and $\;g\;$ is onto.
So the correct answer is $(2)$.

Addendum .
Claim 1 :$\quad g\circ f\;$ is one-one $\implies f\;$ is one-one.
Proof : $\;$ Let $\;a_1\;$ and $\;a_2\;$ be two elements of the set $\;A\;.$
$f(a_1)=f(a_2)\;$ implies that $\;(g\circ f)(a_1)=(g\circ f)(a_2)\;$ and, since $\;g\circ f\;$ is one-one, we get that $\;a_1=a_2\;.$
So we have proved that $f(a_1)=f(a_2)\;$ implies $\;a_1=a_2\;,\;$ consequently $\;f\;$ is one-one.

Claim 2 :$\quad g\circ f\;$ is onto $\implies g\;$ is onto.
Proof : $\;$ Since $\;g\circ f\;$ is onto, it follows that
for any $\;c\in C\;$ there exists $\;a\in A\;$ such that $\;(g\circ f)(a)=c\;,$
consequently,
for any $\;c\in C\;$ there exists $\;b=f(a)\in B\;$ such that $\;g(b)=(g\circ f)(a)=c\;,$
hence $\;g\;$ is onto.
