Proposition 1   For an arbitrary $f$ , $f\perp g$ for any isomorphism $g$ . Dually, if $g$ is arbitrary, then $f\perp g$ for any isomorphism $f$ .

Proof. Suppose $x\circ f = g\circ y$ , and $z\circ g=1_C$ and $g\circ z=1_D$ . Then by defining $h:=z \circ x$ , we get $h\circ f= z\circ x \circ f= z \circ g\circ y = 1_C \circ y = y$ , and $g\circ h= g\circ z\circ x=1_D \circ x = x$ . This shows the existence of $h$ .

If $h':B\to C$ is another morphism such that $h'\circ f = y$ and $g\circ h'=x$ . Then $h = z\circ x = z \circ g\circ h' = 1_C \circ h' = h'$ , showing that $h$ is unique.

The proof of the dual statement is similar.

Proposition 2   If $f\perp f$ , then $f$ is an isomorphism.

Proof. This follows from the commutative diagram

So we get a ``diagonal'' morphism $g:B\to A$ making the resulting diagram commutative again. So, $f\circ g=1_B$ and $g\circ f=1_A$ , or $f$ is an isomorphism.

Proposition 3   Suppose $(g,h)$ is a composable pair of morphisms ($h\circ g$ exists). If $f\perp g$ and $f\perp h$ , then $f\perp (h\circ g)$ . Similarly, if $g\perp f$ and $h\perp f$ , then $(h\circ g)\perp f$ .

Proof. Suppose $f:A\to B$ , $g:C\to D$ and $h:D\to E$ are morphisms as described above, and have a commutative diagram

Because $f\perp h$ , we get a unique morphism $s:B\to D$ and a commutative diagram

Because $f\perp g$ , we have a unique morphism $t:B\to C$ , so the commutative square I above becomes

Since $(h\circ g)\circ t = h\circ (g\circ t) = h\circ s = y$ , we get a commutative diagram

showing that $f\perp (h\circ g)$ . The dual statement of this is proved similarly.