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The universal way to invert a morphism

If you have an endomorphism \(f : X \to X\), you can construct the colimit of the infinite sequence \(X \xrightarrow{f} X \xrightarrow{f} X \xrightarrow{f} \cdots\) to invert \(f\). A familiar example of this comes from algebra: say \(R\) is a commutative ring and \(M\) is an \(R\)-module. Then if \(r \in R\) is some element, the colimit of the sequence \(M \xrightarrow{f} M \xrightarrow{f} \cdots\), where \(f\) is multiplication by \(r\), is simply \(M[r^{-1}] = R[r^{-1}] \otimes_R M\). Of course, in this case, there are easier ways to construct \(M[r^{-1}]\) than taking that colimit, but in more general categories, the sequential colimit recipe might be all you have.

Before we move on, let’s quickly check the claim for \(R\)-modules. There is a cocone over the sequential diagram with vertex \(M[r^{-1}]\) and structure maps \(x \mapsto x/r^n\) on the \(n\)-th copy of \(M\). Now say you have a sequence of maps \(f_i : M \to N\) such that \(f_{i+1}(r x) = f_i(x)\) for all \(i\). Then we can define \(f : M[r^{-1}] \to N\) as \(f(x/r^n) = f_n(x)\). It is easy to check this is well-defined and is the unique map factoring the cocone with vertex \(N\) through the cocone with vertex \(M[r^{-1}]\).

I’ve seen this sequential colimit construction used many times (the homotopy colimit version is very popular in homotopy theory), without really thinking about the sense in which this is the universal way to invert a morphism, but it always seemed a little arbitrary to me: Why a sequential colimit? Why not the colimit of a diagram shaped like a loop? Why a colimit at all? Why not a limit? I finally decided to work it out and the answer is actually quite nice! I’ll explain after setting up a bit of notation.

Let \({\mathbb{N}}\) be the monoid of natural numbers under addition, and let \({B{\mathbb{N}}}\) be the one-object category corresponding to it. Then we can define the category of endomorphisms of objects of a category \({\mathcal{C}}\) to be simply the functor category \({\mathrm{Fun}}({B{\mathbb{N}}}, {\mathcal{C}})\). An object of this category is an object \(X\) of \({\mathcal{C}}\) together with an endormorphism \(f : X \to X\); a morphisms from \((X,f)\) to \((Y,g)\) is simply a morphism \(h : X \to Y\) such that \(h \circ f = h \circ g\). Similarly, we can say that the category of automorphisms of objects of \({\mathcal{C}}\) is given by \({\mathrm{Fun}}({B{\mathbb{Z}}},{\mathcal{C}})\). There is a forgetful functor from automorphisms to endomorphisms which simply forgets the inverse of an automorphism, in terms of these functor categories it is given by composing with the inclusion \({B{\mathbb{N}}}\to {B{\mathbb{Z}}}\). Now we can say in which sense the sequential colimit universally inverts a morphism:

If \({\mathcal{C}}\) is cocomplete, the forgetful functor \({\mathrm{Fun}}({B{\mathbb{Z}}},{\mathcal{C}}) \to {\mathrm{Fun}}({B{\mathbb{N}}},{\mathcal{C}})\) has a left adjoint whose value \((Y,g)\) on \((X,f)\) is given by \(Y = {\mathop{\mathrm{colim}}}\left( X \xrightarrow{f} X \xrightarrow{f} \cdots \right)\) and \(g : Y \to Y\) the morphism induced on the colimit by \(f\).

The left adjoint is given by the left Kan extension along the inclusion \(i : {B{\mathbb{N}}}\to {B{\mathbb{Z}}}\), which we can compute by the usual pointwise formula as a colimit over the comma category \(i \downarrow \bullet\), where \(\bullet\) is the lone object of \({B{\mathbb{Z}}}\). This comma category is easily seen to be the poset \(({\mathbb{Z}}, \ge)\), and the diagram whose colimit gives the Kan extension is easily seen to be \(\cdots \xleftarrow{f} X \xleftarrow{f} X \xleftarrow{f} \cdots\), which has the singly-infinite diagram as a cofinal piece.

Exercise. Why doesn’t this same argument prove that the left adjoint to the identity functor on \({\mathrm{Fun}}({B{\mathbb{N}}}, {\mathcal{C}})\) or on \({\mathrm{Fun}}({B{\mathbb{Z}}}, {\mathcal{C}})\) inverts morphisms? (Obviously the left adjoint to the identity is again the identity, but what step of the argument fails for it?)

Finally, with this new understanding we can answer my earlier questions about other shapes of diagrams you can build out of a single endomorphism \(f : X \to X\):

Omar Antolín Camarena