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It is a separate question to my previous one.

The problem requires me to show that the following statement is true or false.

Let $A,B$ be sets, and $f:A\to B$, $g:B\to A$ be functions. Suppose $g\circ f\circ g$ is surjective, and $f\circ g\circ f$ is injective. Then $f\circ g$ is bijective.

So basically

“$g\circ f\circ g$ is surjective and $f\circ g\circ f$ is injective“ $\implies$ “$g\circ f$ is bijective“ $\implies$“$g$ is surjective and $f$ is injective”

But since it does not imply that “$f\circ g$ is bijective”, can I simply conclude that “$f\circ g$ is not bijective”?

鈴木悠真
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    Failing to proof something is different from prooving that something if false. – Lelouch May 07 '22 at 13:17
  • Okay so I have to find some counter examples. – 鈴木悠真 May 07 '22 at 13:29
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    I always try to start with the simplest possible example to gain intuition. Try $A={0}$ and $B={0,1}$, and see if you can find $f$ and $g$ that satisfy the two conditions. – Joe May 07 '22 at 16:29
  • Okay I will try. Thank you for your suggestion. – 鈴木悠真 May 07 '22 at 16:30
  • Oh wait but $f:{0}\to {0,1}$ forces $f(0)=0$ and $f(0)=1$ which violates the definition of function. – 鈴木悠真 May 07 '22 at 16:40
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    For the $A$ and $B$ that I suggested, there are only two possible choices of $f$, given by either $f(0)=0$ or $f(0)=1$. Also, with those $A$ and $B$, there is only one possible function $g: B \to A$. If either $f$, when combined with $g$, satisfies the two conditions, then you'll have a counter example. – Joe May 07 '22 at 17:26
  • But if we only take one of the case of $f(0)$, say, $f(0)=0$, then can we say that the range of $f$ is still ${0,1}$? As it is impossible that $x\in A$ and $f(x)=1$. – 鈴木悠真 May 07 '22 at 18:38
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    "range" is unfortunately a term that can have one of two possible meanings: the "codomain", which for $f$ is $B$, or the "image" $f(A)$, which is the subset of $B$ such that there are actually points in $A$ that map to them. If a function is surjective, then the "image" is the "codomain". When they are not the same, which one is meant by "range" depends on the source, although perhaps one meaning is more common than the other. If so, I'm not sure. – Joe May 07 '22 at 19:12
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    But, just to be clear, $f: {0} \to {0,1}$ defined by $f(0)=0$ *is* a function, i.e. a relation (subset of $A \times B$) that is both "left-total" and "right-unique". – Joe May 07 '22 at 19:16
  • I understand it now. Thanks for your brief explanation! – 鈴木悠真 May 07 '22 at 19:25
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    You're welcome. – Joe May 07 '22 at 19:44

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