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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

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  • $\begingroup$ The correct answer is $(2)$. $\endgroup$
    – Antonio
    Jul 29, 2021 at 15:27
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    $\begingroup$ Maverick, look at my answer. Do you manage to prove the two claims I wrote ? If you want, I will write the proofs. $\endgroup$
    – Antonio
    Jul 29, 2021 at 15:36

2 Answers 2

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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.

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  • $\begingroup$ What if you take $A=B=C=\mathbb{R}$ and $f(x)=g(x)=e^x$? $\endgroup$
    – Matthew H.
    Jul 29, 2021 at 17:49
  • $\begingroup$ @MatthewPilling, if $A=B=C=\mathbb R$ and $f(x)=g(x)=e^x$, then $(g\circ f)(x)=e^{e^x}$ is one-one but it is not onto, so there does not exist $(g\circ f)^{-1}$. Moreover, since $(g\circ f)$ is one-one, then $\;f\;$ is one-one too. $\endgroup$
    – Antonio
    Jul 29, 2021 at 18:26
  • $\begingroup$ In my example $(g\circ f)^{-1}(x)$ does exist. It's equal to $\ln(\ln(x))$. The problem doesn't indicate that the domain of $(g\circ f)^{-1}$ is all of $C$. In general, the domain of $(g\circ f)^{-1}$ if it exists is equal to $(g\circ f)(A)$ which is a subset of $C$. In my example the domain of $(g\circ f)^{-1}$ is $ (1,\infty)$. You should note that $(g\circ f)^{-1}$ exists but $g$ is not surjective. $\endgroup$
    – Matthew H.
    Jul 29, 2021 at 18:39
  • $\begingroup$ $(g\circ f)^{-1}$ exists if $\;C=(1,+\infty)$. In this case, $\;f\;$ is one-one and $\;g\;$ is onto. See the following link: en.wikipedia.org/wiki/Inverse_function#Definitions $\endgroup$
    – Antonio
    Jul 29, 2021 at 18:41
  • $\begingroup$ Actually, $\;\ln\big(\ln(x)\big):(1,+\infty)\to\mathbb R\;$ is the inverse function of $\;g\circ f:\mathbb R\to(1,+\infty)\;$, whereas the function $\;g\circ f:\mathbb R\to\mathbb R\;$ is not invertible because it is only one-one but not onto. $\endgroup$
    – Antonio
    Jul 29, 2021 at 18:53
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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.

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    $\begingroup$ The correct answer is $(2)$. $\endgroup$
    – Antonio
    Jul 29, 2021 at 15:27

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