Is the inverse limit of quotient the quotient of inverse limits?

360 Views Asked by At

I am recently studying inverse limits and I have the following issue :

Le $V$ be a vector space and $W \subset V$ a subspace of $V$.

Assume that $V$ is equipped with a decreasing filtration $(\mathcal{F}^m V)_{m \in \mathbb{N}}$. We consider $\hat{V}$ the completion of $V$ for this filtration : the inverse limit of the inverse system $(V/\mathcal{F}^m V, V/\mathcal{F}^{m+1} V \rightarrow V/\mathcal{F}^m V)$. i.e. $\hat{V}:=\underset{\longleftarrow}{\lim} V/\mathcal{F}^m V$.

The subspace $W$ is naturally equipped with the filtration $\mathcal{F}^m W:= W \cap \mathcal{F}^m V$. Therefore, we define $\hat{W}$ to be the completion of $W$ for this filtration : $\hat{W}:=\underset{\longleftarrow}{\lim} W/ \mathcal{F}^m W$.

Now, let us consider the quotient space $V/W$ and the canonical projection $\pi : V \rightarrow V/W$.

The space $V/W$ is equipped with the filtration $\pi(\mathcal{F}^m V)$. We define $\hat{V/W}$ to be the completion of $V/W$ for this filtration : $\hat{V/W} := \underset{\longleftarrow}{\lim} (V/W)/\pi(\mathcal{F}^m V)$.

Question : Do we have $\widehat{V/W} \simeq \hat{V}/\hat{W}$ ?

This is equivalent to $\underset{\longleftarrow}{\lim} (V/W) \big/ \pi (\mathcal{F}^m V) = (\underset{\longleftarrow}{\lim} V/\mathcal{F}^m V) \Big/ \underset{\longleftarrow}{\lim} W / (W \cap \mathcal{F}^m V)$.

Intuitively, it seems true since the filtrations on $W$ and on $V/W$ are naturally defined by the canonical inclusion and canonical projection respectively. But I do not have a clue whether it is true or false. If true my first thought is to begin by $$(V/W) \big/ \pi(\mathcal{F}^m V)=(V/W) \big/ (\mathcal{F}^m V/\mathcal{F}^m V \cap W) \simeq (V/\mathcal{F}^m V) \big/ (W/\mathcal{F}^m V \cap W).$$ But I do not know if the last identification is true for any $m \in \mathbb{N}$ and if it is the case how to introduce the inverse limit ?

If false, is there any counter-example for this ?

Any hint or help would be appreciated !

Thanks in advance.