Hi I am trying to formulate a proof for this proposition:

Let $A$ and $B$ be sets and $f$ a map such that $f:A \to B$

$f$ has a right inverse $\implies$ $f$ is surjective


The statement $f$ has a right inverse $\implies$ $\exists$ a function $g:B\to A$

such that $f\circ g(b) = id_B$ $\forall b \in B$

I'm concerned about my logic here:

"This statement implies that every element of $B$ lies in the pre-image of $f$

thus $f$ is surjective as $\forall b \in B$ $\exists$ $a \in A $ such that $f(a) = b$"

I feel this logic is not water tight and don't know how to formulate it.



You're really there already I would say, but to write it out fully:

Let $b \in B$. Our goal is to prove that it has a pre-image under $f$. Notice that $f(g(b)) = b$, since $f \circ g = \text{Id}_B$. Therefore $g(b)$ is a pre-image for $b$ under $f$, because $f$ maps $g(b)$ to $b$.


You can obtain it with part of a very general result:

Let $A, B, C$ be sets and $\;f:A\longrightarrow B$, $\;g:B\longrightarrow C$. Then

  • If $g\circ f$ is injective, $f$ is injective.

  • If $g\circ f$ is surjective, $g$ is surjective.


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