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As a complement (no pun intended) to Zhen Lin's excellent answer, here is a more direct, diagrammatic way to see that $R'$ is symmetric. Consider the following diagram:

Both $s$ and $\delta$ are involutions, so the bottom left square is a pullback; the top left square is a pullback by disjointness of coproducts and the right rectangle is a pullback by construction. Hence by stability of coproducts under pullback, the top row is a coproduct diagram, which means that $\pi_1$ is invertible, which in turn means that $\delta \circ r'$ factors through $r'$ via $\pi_2$.

As a complement (no pun intended) to Zhen Lin's excellent answer, here is a more direct, diagrammatic way to see that $R'$ is symmetric. Consider the following diagram:

Both $s$ and $\delta$ are involutions, so the bottom left square is a pullback; the top left square is a pullback by disjointness of coproducts and the right rectangle is a pullback by construction. Hence by stability of coproducts under pullback, the top row is a coproduct diagram, which means that $\pi_1$ is invertible, which in turn means that $\delta \circ r'$ factors through $r'$ via $\pi_2$.

As a complement (no pun intended) to Zhen Lin's excellent answer, here is a more direct, diagrammatic way to see that $R'$ is symmetric. Consider the following diagram:

Both $s$ and $\delta$ are involutions, so the bottom left square is a pullback; the top left square is a pullback by disjointness of coproducts and the right rectangle is a pullback by construction. Hence by stability of coproducts under pullback, the top row is a coproduct diagram, which means that $\pi_1$ is invertible, which in turn means that $\delta \circ r'$ factors through $r'$ via $\pi_2$.

As a complement (no pun intended) to Zhen Lin's excellent answer, here is a more direct, diagrammatic way to see that $R'$ is symmetric. Consider the following diagram:

Both $s$ and $\delta$ are involutions, so the bottom left square is a pullback; the top left square is a pullback by disjointness of coproducts and the right rectangle is a pullback by construction. Hence by stability of coproducts under pullback, the top row is a coproduct diagram, which means that $\pi_1$ is invertible, which in turn means that $\delta \circ r'$ factors through $r'$ via $\pi_2$.

As a complement (no pun intended) to Zhen Lin's excellent answer, here is a more direct, diagrammatic way to see that $R'$ is symmetric. Consider the following diagram:

Both $s$ and $\delta$ are involutions, so the bottom left square is a pullback; the top left square is a pullback by disjointness of coproducts and the right rectangle is a pullback by construction. Hence by stability of coproducts under pullback, the top row is a coproduct diagram, which means that $\pi_1$ is invertible, which in turn means that $\delta \circ r'$ factors through $r'$ via $\pi_2$.

As a complement (no pun intended) to Zhen Lin's excellent answer, here is a more direct, diagrammatic way to see that $R'$ is symmetric. Consider the following diagram:

Both $s$ and $\delta$ are involutions, so the bottom left square is a pullback; the top left square is a pullback by disjointness of coproducts and the right rectangle is a pullback by construction. Hence by stability of coproducts under pullback, the top row is a coproduct diagram, which means that $\pi_1$ is invertible, which in turn means that $\delta \circ r'$ factors through $r'$ via $\pi_2$.