$\exists h: B \rightarrow A. f \circ h = id_B$ **NB**: If $f$ is bijective and $g \circ f = id_A$ and $f \circ h = id_B$ then $h=g$ $g \circ f \circ h$: $g \circ f ...
Some results have been hidden because they may be inaccessible to you
Show inaccessible results
Feedback