Orthogonality relation between morphisms in a category is not symmetric In Borceux' Handbook of Categorical Algebra (Dfn 5.4.1), the authors introduces the notion of orthogonality between morphisms and asserts that such a relation is not symmetric.
Recall: f is orthogonal to g if whenever there is a commutative diagram $u \circ f=g \circ v$, then there exists a unique diagonal d such that $v=d \circ f$ and $g \circ d=u$.
I tried to provide some counterexamples to symmetry. For instance, in the category ${\bf Set}$ I took an epimorphism $f$ and a monomorphism $g$. It is well known that $f \perp g$. The converse does not hold in general, as one may see by taking the monomorphism $g: X \to Y$, with $X=\{x\}$ a singleton, $Y=\{y_1,y_2\}$ and $g(x)=y_1$, an epimorphism $f: Z \to W$, with $Z=\{z_1,z_2,z_3\}$, $W=\{w_1,w_2\}$ and $f(z_1)=f(z_2)=w_1$ and $f(z_3)=w_2$. If we consider the maps $h: Y \to W$ with $h(y_1)=h(y_2)=w_1$ and $l: X \to Z$ with $l(x)=w_1$, we clearly have $h \circ g=f \circ l$. However, I have to possible diagonals, depending on the choice of the image of $y_2$ between $z_1,z_2$.
Well, is there a simple example in ${\bf Set}$ (or in some manageable category) where the symmetry of the orthogonality relation fails because of the non-existence of the diagonal?
 A: The simplest example for orthogonality failing to be symmetric is the finite category on four objects consisting of a commutative square $u\circ f=g\circ v$. Indeed, $f\perp g$ does not hold because there is no diagonal, while $g\perp f$ holds vacuously as there is no commutative diagram $u'\circ g=f\circ v'$.
Similarly, the finite category consisting of a commutative square and several diagonals has $f\perp g$ not holding due to the multiple diagonals, but $g\perp f$ holding vacuously.
An example in Set with monomorphisms and epimorphisms with no diagonals does not exists unless the axiom of choice fails. Indeed, suppose $u\circ g\colon X\hookrightarrow Y\to W$ is the same as $f\circ v\colon X\to Z\to W$ for $g\colon X\hookrightarrow Y$ a monomorphism, i.e. an injective function. We have an isomorphism $Y=X\sqcup A$ with a disjoint union. The condition on the diagonal $d$ that $d\circ g=v$ is then the condition that $d|_X=v$, in which case the condition that $f\circ d=u$ reduces to $f\circ d|_A=u|_A$ (since $f\circ d|_X=f\circ v=u\circ g=u|_X$).
Since $d$ is completely determined by its restrictions to $X$ and $A$, the only way it would not exist is there was no $d|_A\colon A\to Z$ for which $u|_A\colon A\to W$ would factor as $u|_A=f\circ d|_A$.
In other words, any example where symmetry of orthogonality fails due to lack of a diagonal amounts to $u\circ A\to W$ not factoring through $f\colon Z\to W$. But if $f\colon Z\twoheadrightarrow W$ is an epimorphism, i.e. a surjective function, such a factorization corresponds to a choice of an element in $f^{-1}(u(a))$ for each element $a\in A$. Thus the lack of existence of such a factorization corresponds to a failure of the axiom of choice.
