if $C$ is a filtered coalgebra, does Gr($B\Omega C)\backsimeq B\Omega ($Gr $C)$ hold?

Question

I have heard that under some assumptions, the functor 'Gr' from filtered graded objects with exhaustive filtration to graded objects $X\rightarrow$ Gr$(X)$ commutes with direct sums (this seems to be clear) and tensor products. If so, the question on the title is solved. Can anybody please tell me an example of such assumptions so that this properties are true?. Above all for tensor products, I mean, when does Gr$(X^{\otimes j})\backsimeq$ $($Gr$(X))^{\otimes j}$ hold?.

Answer 1

The claim is obvious if you are working with vector spaces: If you have a filtered object $X = \bigcup X_{\leq n}$, then you can choose a basis of $X_{\leq 0}$, extend it to a basis of $X_{\leq 1}$, extend to a basis of $X_{\leq 2}$, and so on. Then this basis provides you a filtered isomorphism $X \cong \mathrm{gr}X$, and moreover provides an isomorphism $\mathrm{gr}(X\otimes Y) \cong \mathrm{gr}X \otimes \mathrm{gr}Y$. Finally, you can check directly that the latter isomorphism actually doesn't depend on the choice of basis, and so is a natural isomorphism, and makes $\mathrm{gr}$ into a symmetric monoidal functor.

If you don't have access to bases, the claim can fail. Consider the category of filtered abelian groups. Set $X = \mathbb{Z}/4$, with its filtration $X_{\leq 0} = \{0,2\} \subset X_{\leq 1} = X = \{0,1,2,3\}$. Then $X \otimes X \cong \mathbb{Z}/4$ with some filtration. But $\mathrm{gr}X = \mathbb{Z}/2 \oplus \mathbb{Z}/2$, with the first summand in degree $0$ and the second summand in degree $1$, and so $\mathrm{gr}X \otimes \mathrm{gr}X = \mathbb{Z}/2 \oplus (\mathbb{Z}/2)^2 \oplus \mathbb{Z}/2$, supported in degrees $0,1,2$.