9

The associated graded functor has an obvious universal property if you use a sufficiently nice definition of the notion of "being filtered". A good notion of the category of filtered objects over a category $\mathcal{C}$ consists of the functor category $\text{Fun}((\mathbb{Z},\leq), \mathcal{C})$, where the poset $(\mathbb{Z},\leq)$ is viewed as a category. ...

7

Your argument has an error: $\sigma_B = t$ does not imply $X_t \in B$, only $X_t \in \overline{B}$.
Consider the following example. Let $\mathcal{X}$ be the real line and let $Z$ be a fair coin flip, so $Z = \pm 1$ with probability $1/2$. Set $X_t = tZ$. So this process flips a coin at time 0 to decide whether to move left or right, then moves in that ...

3

This is proved in Proposition 4.4 of "Deconstructibility and the Hill Lemma in Grothendieck categories" by Jan Šťovíček.

answered Jun 30 '15 at 11:33

Jeremy Rickard

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3

Yes, first one should check that the restriction of a covering space is a covering space. This is true either by just checking the axioms or appealing to the fact that the pullback of a covering space is a covering space, and the restriction of a covering space is just the pullback of an inclusion into the base space.
With that settled, the standard CW ...

2

Yes, if $A = F^0A$ then $1 \in F^0A$ so $f(1) = 1 \in F^0 B$ and so the ideal $F^0 B$ contains all of $B$. If you do not require unital dgas/maps, then the answer is no, just take the zero map and any exhaustive filtration at the source.
No, consider $A = k[x]$, $F^n A = (x^n)$, $B = k$, and $f : A \to B$, $f(x) = 1$. Then $\bigcap_n F^n A = 0$ but $\...

2

There is recent work on the homotopical algebra of the simplicial analoque of your situation. Lurie defined a P-stratified space as being a space over the Alexandroff space corresponding to P. Very recently in his thesis,
https://arxiv.org/abs/1908.01366,
Sylvain Douteau has looked at this from the point of view of simplicial sets over the nerve of P. ...

1

As I said in comments, there is a fair amount of literature to be found by Googling "infinite socle series".
More specifically, a module $M$ for which (in the notation of the question) $\overline{\text{soc}}(M)=M$ is called a "semi-artinian module", and a ring $R$ for which every module is semi-artinian (or equivalently for which $R$ is semi-artinian as a ...

answered Jun 13 '20 at 18:14

Jeremy Rickard

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1

Yes, at least if $X_0=0$. The containment $\mathfrak G_t\subset\mathfrak F_t$ is clear. In the other direction, is $0<s\le t$ then $X_s=(H\cdot X)_t$, where $H_u(\omega) :=1_{]0,s]}(u)$ is (trivially) predictable. This shows that $X_s$ is $\mathfrak G_t$-mesurable for each $s\in]0,t]$.

1

I always find it helpful to write down the unit and counit in order to understand an adjunction, so I'll just expand on Nicolas Schmidt's excellent answer.
From the point of view discussed by Nicolas, let us consider filtered and graded objects in an abelian category. Denote by $\operatorname{triv}$ the trivial (generalized) filtration, so that the ...

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