Let $\text{Arr}(\mathbb{C})$ denote the set of arrows of small category $\mathbb{C}.$ Let $M \subseteq \text{Arr}(\mathbb{C})$ be the set of split monomorphisms of $\mathbb{C}.$ I want to choose a retraction $f^{-1}$ for each $f \in M$ with the feature that whenever I take a composable pair of arrows $A \overset{f}\to B$ and $B \overset{g}\to C$ I have that the composition $gf$ has chosen retraction $(gf)^{-1}$ equal to the composition $f^{-1} g^{-1}.$
What is required of $\mathbb{C}$ to make this possible ? Is it always possible ?
Let $\mathbb{C}_M$ be the wide subcategory of $\mathbb{C}$ on the split monomorphisms. I guess my request is equivalent to asking for an identity on objects functor $\mathbb{C}_M$ to $\mathbb{C}^{op}$ that sends each arrow of $M$ to one of its retractions, but I'm not sure.
This is not possible for the category of finite sets.
First note that for $f$ an isomorphism, the only possible choice for $f^{-1}$ is the usual inverse of $f$ (so the notation doesn't conflict).
Let $X=\{1,2\}$ and $Y=\{0,1,2\}$.
Let $\sigma:X\to X$ be the transposition that swaps $1$ and $2$, so $\sigma^{-1}=\sigma$.
Let $\tau:Y\to Y$ be the transposition that swaps $1$ and $2$ and fixes $0$, so $\tau^{-1}=\tau$.
Let $f:X\to Y$ be the inclusion map. There are two possible choices for $f^{-1}$. Without loss of generality choose the one that sends $0$ to $1$.
Then $f\sigma=\tau f$, so $(f\sigma)^{-1}=(\tau f)^{-1}$. But $\sigma^{-1}f^{-1}\neq f^{-1}\tau^{-1}$ since $\sigma^{-1}f^{-1}$ sends $0$ to $2$ but $f^{-1}\tau^{-1}$ sends $0$ to $1$.