Suppose outer square in above diagram is comutative i.e., $c\circ q\circ p=s\circ r\circ a$
Further, suppose right side square is commutative i.e., $c\circ q=s\circ b$?
Does it imply left side square is commutative i.e., $b\circ p=r\circ a$??
If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?
Best Answer
In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the left-hand square commutes, since $$s \circ r \circ a = c \circ q \circ p = s \circ b \circ p \quad \Rightarrow \quad r \circ a = b \circ p$$
In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.
So let $f,g : B \to E$ and suppose $s \circ f = s \circ g$. Form the diagram with $a=p=q=\mathrm{id}_B$, $b=f$, $r=g$ and $c = s \circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s \circ f = s \circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.