# Counit for the filtration adjunction

This is a follow-up to my question here, which was answered by Eric Wofsey.

Let $\mathsf{A}$ be an abelian category, such as $R \mathsf{Mod}$ for concreteness. We can think of the category of $\mathbb{Z}$-graded objects in $\mathsf{A}$ as the functor category $\mathsf{A}^\mathbb{Z}$ where $\mathbb{Z}$ is viewed as a discrete category. This works for $\mathbb{N}$-gradings as well.

The category of filtered objects in $\mathsf{A}$ in the classical sense may (potentially) be too small, so we can allow "generalized" filtrations as Nicolas Schmidt suggested in his answer to a related question. Then the generalized filtrations are the objects of the functor category $\mathsf{A}^{(\mathbb{Z}, \leq)}$, where $(\mathbb{Z}, \leq)$ is the ordered group of integers viewed as a poset category. We could of course use the poset $(\mathbb{N}, \leq)$ instead if we need to.

Now, given an a graded object $A$ we can form the corresponding filtered object with pieces $A^{(n)} = \bigoplus_{i \leq n} A_i$, where the arrows between the pieces are just inclusions $\bigoplus_{i \leq {n-1}} A_i \hookrightarrow \bigoplus_{i \leq n} A_i$. This is obviously a functor from graded objects in $\mathsf{A}$ to filtered objects in $\mathsf{A}$. Indeed, given a morphism $f: A \to B$ of graded objects, which we write in degree $i$ as $f_i: A_i \to B_i$, the corresponding morphism of filtered objects is the obvious one $f^{(n)} := \bigoplus_{i \leq n} f_i : A^{(n)} \to B^{(n)}$.

As pointed out by Eric Wofsey, this filtration functor $F$ should be left adjoint to the "forgetful functor" $G$ from generalized filtrations to gradings. He argued by Kan extensions, but unfortunately I am not familiar with Kan extensions so I tried to prove it the old-fashioned way: by coming up with the unit and counit for the adjunction. However, I ran into an issue: what should the counit be?

Suppose $a_i: A_{j-1} \to A_j$ is a "generalized" filtered object. The counit should have components being filtered maps $\varepsilon_A: FG(A) \to A$. The filtered object $FG(A)$ has the obvious inclusions $\iota_j: \bigoplus_{i \leq j-1} A_i \hookrightarrow \bigoplus_{i \leq j} A_i$, so on the $j$th filtered piece the counit is a map $(\varepsilon_A)_j : \bigoplus_{i \leq j} A_i \to A_j$ which must satisfy $$(\varepsilon_A)_j \circ \iota_j = a_j \circ (\varepsilon_A)_{j-1}.$$

But if we take $R \mathsf{Mod}$ for our abelian category, then on an element $(\ldots, x_{j-2}, x_{j-1}) \in \bigoplus_{i \leq j-1} A_i$, we must have $$(\varepsilon_A)_j(\ldots, x_{j-2}, x_{j-1}, 0) = a_j((\varepsilon_A)_{j-1}(\ldots, x_{j-2}, x_{j-1}))$$ So, the obvious choice of counit as the projection onto the $j$th component doesn't work unless $a_j$ is an inclusion map. This makes me wonder if maybe the adjunction only holds when we restrict to classical filtrations where the maps between the pieces are inclusions.

The counit isn't just the projection. For each $i\leq j$, we have a map $a_{ij}:A_i\to A_j$ obtained by composing $j-i$ of the $a$ maps. You then want to define $(\epsilon_A)_j(\dots,x_{j-2},x_{j-1},x_j)=\sum_{i\leq j} a_{ij}(x_i)$.
I find the adjunction easiest to think of in terms of a Hom-set bijection. Given a map $FA\to B$, you obtain a map $A\to GB$ by defining $A_i\to B_i$ to be the restriction of the map $(FA)_i\to B$ to the summand $A_i\subset (FA)_i$. Conversely, given $A\to GB$, you define $FA\to B$ by saying the map $(FA)_i\to B$ is given on the $A_j$ summand by the map $A_j\to B_j$ composed with the map $B_j\to B_i$ given by the filtration on $B$. You can check that this is the unique way you can define $FA\to B$ which is both compatible with the filtration maps and restricts to the map $A_i\to B_i$ on the $A_i$ summand of each $(FA)_i$.
You can then check that the definition of $\epsilon_A$ I wrote above is just what you get when you apply this Hom-set bijection to the identity map $GA\to GA$.