Inverse limit and graded functor commute?

I am trying to understand a proof where there are graded algebras and inverse limit involved.

In one of the steps it seems to commute this two elements. Is there any reference where this is stated.

$$\varprojlim (gr(\Lambda^*_n))=gr(\varprojlim \Lambda_n^*).$$

Where $$\Lambda^*_n$$ stands for the shifted symmetric polynomials of $$n$$ indeterminates.

Note that the inverse limit $$\varprojlim \Lambda_n^* = \Lambda^*$$ is taken in the category of filtered algebras.

MSE QUESTION HERE.

• This seems rather similar to your question on Mathematics: inverse limit and graded functor commute?. I think that this answer provides reasonable guidelines on cross-posting, probably one of the most important things is to link each copy to the other ones. – Martin Sleziak Jun 7 at 18:40
• Oh! Hate when people is more worried about format than Maths. I do it to get more people involved. Anything else. I think that is not a big trouble to post it in both sites. Spending the time on looking if I post it in other site or not I think is not what people should do when they join a MATHS EXCHANGE site. – idriskameni Jun 7 at 18:46
• By "shifted polynomial algebra" you mean the exterior algebra? – Marco Farinati Jun 11 at 19:18
• Filtered by degree or by order? – Marco Farinati Jun 11 at 19:19
• @MarcoFarinati I define $\Lambda^*$ as the inverse limit of $\Lambda_n^*$ in the category of filtered algebras. It is graded by degree, i.e. $\Lambda_n^*=\cup_k (\Lambda_n^*)^k$. Here, each $(\Lambda_n^*)^k$ is the algebra containing all shifted symmetric polynomials in $n$ indeterminates and with terms with degree lower or equal to $k$. – idriskameni Jun 11 at 19:51