Proof.  [#242]

Let \(f : b \to a\) be a vertical morphism, i.e. \(p(f) = \textsf {id}_{x}\). Then the factorization \(\textsf {id}_{x} = \textsf {id}_{x} \circ \textsf {id}_{x}\) lifts to a unique pair \(\alpha \) and \(\beta \) over \(\textsf {id}_{x}\) with \(\alpha \circ \beta = f\). Since \(f = f \circ \textsf {id}_{b} = \textsf {id}_{a} \circ f\) we must have \(a=b\), \(\alpha = \textsf {id}_{a}\) and \(\beta = \textsf {id}_{a}\) and so \(f = \textsf {id}_{a}\).