$g\circ f$ bijective iff $f$ and $g$ bijective?

Question

Is the following true:

$g\circ f$ bijective iff $f$ and $g$ bijective?

Or can the requirements be weakened for $g$ (i.e. $g$ only injective or surjective)? Or $f$?

Answer 1

Hint: one direction is true. For the other, consider $A=\{0\}$, $B=\{1,2\}$, $C=\{3\}$. Define $f\colon A\to B$ by $f(0)=1$; define $g\colon B\to C$ by $g(1)=3$ and $g(2)=3$. Is $g\circ f$ bijective? Is $f$ bijective? Is $g$ bijective?

Answer 2

Let $x=(x_1,x_2,...)$ be a real sequence and define $Rx = (0,x_1,x_2,...)$ and $Lx = (x_2,x_3,...)$. Then $L\circ R $ is the identity but $R \circ L$ is neither injective nor surjective.

It is clear that if $f,g$ are both bijective then so is $f \circ g$.

Answer 3

Let $f:A\rightarrow B$ and $g:B\rightarrow C$

If $g\circ f $ is injective then $f$ is injective.

If $g\circ f$ is surjective then $g$ is surjective.

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$g\circ f$ is injective if and only if $f$ is injective and $g_{f(A)}$ is injective.

Answer 4

1) $f$ and $g$ bijective $\implies$ $f \circ g$ is bijective.

2) $f \circ g$ is bijective $\implies$ [$g$ is injective AND $f$ is surjective]

3) $f \circ g$ is bijective $\implies$ [$g$ is surjective $\iff$ $f$ is injective](but neither need be true.)

Proof:

1a)$x \ne y \implies g(x) \ne g(y) \implies f(g(x)) \ne f(g(x))$ so injective.

1b)$z \implies$ there is a $y$ such that $f(y) = z$; $y \implies$ there is an $x$ such that $y = g(x)$; and therefore $z = f(y) = f(g(x))$ so surjective.

2a) If $x \ne y$ but $g(x) = g(y)$ then $f(g(x)) = f(g(y))$; so $f \circ g$ injective $=> g$ injective.

2b) If there is no $y$ such that $f(y) = z$ then there is no $g(x)$ such that $f(g(x)) = z$; so if $f \circ g$ is surjective, $f$ must be surjective.

3a) If $g$ is surjective then for all $y, v; y \ne v$ there are $x, w; x \ne w$ such that $g(x) = y$ and $g(w) = v$. So $f(y) = f(g(x)) \ne f(g(w)) = f(v)$ so $f$ is injective.

If $g$ is not surjective, however, it would be possible to have an $v, y; y \ne v$ where there is no $x$ such that $g(x) =v$, in which case $f(v) = f(y)$ would not be a contradiction to $f \circ g$ being injective as $f(v) \ne f(g(x))$ for any x.

3b) If $g$ is not surjective then there exist a $y$ such that there is no $x$ such that $g(x) = y$. But there does exist an $x$ such that $f(g(x)) = f(y)$ but $g(x) \ne y$. So $f$ is not injective.

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An example where $f \circ g$ is bijective but $f$ is not injective and $g$ is not surjective.