1
$\begingroup$

Let $\mathcal{C}$ be any category and let $A, B$ be objects. A retraction is a morphism $r: A\to B$ such that there is $s:B\to A$ such that $r\circ s:B\to B$ is the identity.

A morphism $l: A\to B$ is said to be left-factoring if for any $Z\in \mathbf{Ob}(\mathcal{C})$ and any morphism $f: Z\to B$ there is $h: Z\to A$ such that $f = l\circ h$. (I take this definition from Universal and left-factoring order-preserving maps.)

Is there any implication between these terms?

$\endgroup$

1 Answer 1

3
$\begingroup$

Yes - these terms are equivalent.

First, it is easy to see that any left-factoring morphism is a retraction: Suppose $l: A \to B$ and pick $Z:= B$ in your definition above and $f:= \mathsf{id}_B$. Since $l$ is left-factoring there is $h:B\to A$ sich that $\mathsf{id}_B = l\circ h$. So $l$ is a retraction.

Conversely, let $l:A\to B$ be a retraction, let $Z$ be an object and $f:Z\to B$. Because $l$ is a retraction there is $s: B\to A$ such that $l\circ s:B\to B$ is the identity. So we set $h:= s\circ f$ and using associativity of composition we get $l\circ h = l\circ (s\circ f) = (l\circ s)\circ f = \mathsf{id}_B\circ f = f$. So $l$ is left-factoring.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.