Conditions on $f$ and $g$ if $(g\circ f)^{-1}$ exists

Question

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

Answer 1

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.

Answer 2

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.