Suppose f : A ---> B and g : B ---> A are functions for which g o f = 1A...

Question

If I were to suppose that $f : A \to B$ and $g : B \to A$ are functions for which $g \circ f = 1_A$, is $f$ always surjective and is $g$ always injective? How would I either prove this or counter it?

Any help is appreciated.

Answer 1

Take $A = \{a\}$ to be any one-point set, take $B$ to be any set whatsoever, take $f : A \to B$ to be any function whatsoever, and take $g : B \to A$ to be any function whatsoever. Then $g \circ f(a)=a$ and therefore $g \circ f$ is the identity map on $A$. But if $B$ has more than one element then $f$ is not surjective and $g$ is not injective.

Answer 2

Whenever $g\circ f$ is injective, we have that $f$ is injective. For suppose that $f(x)=f(y)$. Then $g(f(x))=g(f(y))$. Since $g\circ f$ is injective, $x=y$.

But, if $g\circ f$ is surjective then only $g$ need be surjective. The hint I'll leave is that you can get a counterexample with finite sets where $A$ is smaller than $B$.