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I know how to show the composition of two surjective functions is surjective or two injective functions is injective, but I don't understand this case problem.

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    $\begingroup$ Is your task to prove it, or is your task to determine whether it is true and then prove your answer? $\endgroup$
    – TStancek
    Sep 4 '17 at 13:10
  • $\begingroup$ Have you looked at examples? I mean, suppose one of the functions is the identity. $\endgroup$
    – lulu
    Sep 4 '17 at 13:12
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That's not true in general.We can take a counterexample to show one case when it is not true:

Let $f:A\to B$ and $g:B\to C$ be two surjective functions and let $h:A\to C$ be their composition such that: $A=\{a,b\}$,$B=\{c,d\}$ and $C=\{e\}$ then

$f(a)=c$; $f(b)=d$ and

$g(c)=e$; $g(d)=e$

It is obvious that $h(a)=e$ and $h(b)=e$ so $h$ is not injective.

So we found an example where the composition of two surjective functions is not injective, therefore your statement is not always true.

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You can't, because it's not true. Here is a counter-example.

Simplifying Linda's answer, take sets $A=\{a,b\}$ and $B=\{q\}$
and constant function $f: A\to B$ and identity function $g:B\to B$.

Then $$g\circ f$$ is a surjection but not an injection.

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