# Extending costandard filtrations?

Let $$\mathsf{C}$$ be a locally finite abelian category. Assume that there exists a poset $$\Lambda$$ and a complete set $$\{L(\lambda)\}_{\lambda \in \Lambda}$$ of representatives of isomorphism classes of simple objects of $$\mathsf{C}$$. Choose such a set $$\Lambda$$ and a corresponding set $$\{L(\lambda)\}_{\lambda \in \Lambda}$$.

We call $$\nabla(\lambda)$$ a costandard object for $$L(\lambda)$$ if $$\operatorname{Soc}\nabla(\lambda)\cong L(\lambda)$$ and all composition factors $$L(\mu)$$ of $$\nabla(\lambda)/\operatorname{Soc}\nabla(\lambda)$$ satisfy $$\mu < \lambda$$. Here, $$\operatorname{Soc}\nabla(\lambda)$$ denotes the socle of $$\nabla(\lambda)$$.

We call $$\Delta(\lambda)$$ a standard object for $$L(\lambda)$$ if $$\operatorname{Hd}\Delta(\lambda)\cong L(\lambda)$$ and all composition factors $$L(\mu)$$ of $$\operatorname{Rad}\Delta(\lambda)$$ satisfy $$\mu < \lambda$$. Here, $$\operatorname{Hd}\Delta(\lambda)$$ denotes the head of $$\Delta(\lambda)$$ and $$\operatorname{Rad}\Delta(\lambda)$$ denotes the radical of $$\Delta(\lambda)$$.

We say that an object $$N \in \mathsf{C}$$ has a costandard filtration if there exists a filtration $$0=N_0 \subsetneq N_1 \subsetneq … \subsetneq N_{n-1} \subsetneq N_n=N$$ by subobjects $$N_i$$ of $$N$$ such that for each $$i$$ the quotient $$N_i/N_{i-1}$$ is isomorphic to a costandard object $$\nabla(\lambda_i)$$, for some $$\lambda_i \in \Lambda$$. Dually, we define a standard filtration.

Question
Assume that $$N \in \mathsf{C}$$ has a costandard filtration. Further assume that $$\tilde{N} \in \mathsf{C}$$ is a subobject of $$N$$ such that $$N/\tilde{N}\cong \nabla (\lambda)$$ for some $$\lambda \in \Lambda$$. Does there exist a costandard filtration of $$N$$ $$0=N_{0}\subsetneq \cdots\subsetneq N_{n-1}\subsetneq N_{n}=N$$ with $$N_{n-1}\cong\tilde{N}$$?

For example, let $$\mathsf{C}$$ be the category of finite dimensional representations of the quiver $$1\rightarrow2$$. Then $$\mathsf{C}$$ has three indecomposable objects: the simple modules $$S(1)$$ and $$S(2)$$ and a two-dimensional module $$X$$ with $$\operatorname{soc}X=S(2)$$ and $$\operatorname{head}X=S(1)$$.
Take $$\nabla(1)=S(1)$$ and $$\nabla(2)=X$$. Then if $$N=X$$ and $$\tilde{N}=\operatorname{soc}X$$, both $$N\cong\nabla(2)$$ and $$N/\tilde{N}\cong\nabla(1)$$ certainly have costandard filtrations, but $$\tilde{N}\cong S(2)$$ does not.
• Does anything change if we require $N$ to be an indecomposable tilting object? I am asking because in this paper arxiv.org/pdf/2107.07887.pdf the authors claim in Lemma 3.1 that $\operatorname{Ker}(\pi^\lambda)$ has a costandard filtration. I was trying to use Remark 2.15 somehow. If the claim of my question had been true, I could have defined $\pi^\lambda$ as the projection of the ultimate element of a certain costandard filtration (whose existence would have been guaranteed by the claim) onto the quotient with the penultimate element. How can I save the argument? Jan 5 at 10:42
• That is, why does $\nabla(\lambda)$ occur at the top of a costandard filtration of $T(\lambda)$? Thank you for your helpful answer by the way. Jan 5 at 10:43
• @Margaret In my example, $N$ is an indecomposable tilting module. ($\Delta(1)=S(1)$, $\Delta(2)=S(2)$, $T(1)=\nabla(1)$ and $T(2)=\nabla(2)$.) Jan 6 at 9:35