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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.

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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 $GA\to GA$.

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  • $\begingroup$ Ah that makes sense! Thanks so much for your help. $\endgroup$
    – ಠ_ಠ
    Jul 30, 2016 at 8:18

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