Truncation of inverse systems

Question

I'm trying to ascertain when (if ever) it is acceptable to ``truncate'' terms of an inverse system and arrive at an isomorphic limit. To simplify matters, assume that our directed system is $\mathbb N$ with the normal ordering and let $A=(A_i,f_{ij})_{0\leq i \leq j}$ be an inverse system. If $\hat A_n = (A_i, f_{ij})_{n\leq i \leq j}$ is the 'truncated' system, when is $\displaystyle \lim_{\longleftarrow}A \cong \lim_{\longleftarrow} \hat A_n$ (assuming the limits exist)?

Examples:

1. [Edit: Made a mistake in my indices]A very simple case is to look in the category of sets and a nesting of sets $X_m \subseteq X_{m-1} \subseteq \cdots \subseteq X_2 \subseteq X_1$. If the morphisms are inclusions, then the inverse limit is the intersection and yields $X_m$. Truncation of the first $n$-terms will still result in an intersection of $X_m$.
2. I'm not certain if this is correct, but I feel that truncation of the inverse system defining the $p$-adics should still result in the $p$-adics. In particular, if $\mathbb Z/\langle p^n\rangle \to \mathbb Z/\langle p^{n-1}\rangle$ is the obvious map then the identification of this inverse system with $\mathbb Z_p$ is done via an analogy of Cauchy sequences in the $p$-adic norm, and it is only the farfield limiting behaviour which really seems to matter, so it seems to me that truncation would not affect this limit.

Any insight would be appreciated.

Answer 1

Let $\Omega=\displaystyle \lim_{\longleftarrow}A$ and $\Omega'=\displaystyle \lim_{\longleftarrow}\hat{A}$. For $i\in\{0,\ldots, n-1\}$ let $p'_i\colon\Omega'\to A_i$ be given by $f_{in}\circ p'_n$ where $p'_n$ is the usual projection of $\Omega'$ onto $A_n$ given by the universal property of limits, and let $p'_j\colon\Omega'\to A_i$ be the usual projection for $j\geq n$. I claim that this set of maps makes $\Omega'$ into a limit of the inverse system $A$ and so is isomorphic to $\Omega$.

We define a map $\lambda\colon \Omega\to\Omega'$ to be the unique map given by the universal property of $\Omega'$ as the limit of the inverse system $\hat{A}$ where the maps $p_i\colon \Omega\to A_i$ for $i\geq n$ are simply the projections.

Let $Y$ be an object in the category and suppose we have maps $\psi_i\colon Y\to A_i$ for each $i\geq 0$ such that $f_{ij}\circ \psi_j=\psi_i$ for all $j\geq i\geq 0$. Well, by the universal property of $\Omega$ as the limit of the inverse system $A$, we have a unique map $\lambda'\colon Y\to \Omega$ such that $p_j\circ\lambda'=\psi_j$ and $p_i\circ\lambda'=\psi_i$.

For $j\geq n$ we have $p'_j\circ\lambda= p_j$ by the definition of $\lambda$ and so $p'_j\circ\lambda\circ\lambda'= p_j\circ\lambda'=\psi_j$.

For $0\leq j\leq n-1$ we have $p'_j\circ\lambda=f_{jn}\circ p'_n\circ\lambda=f_{jn}\circ p_n=p_j$ and so $p'_j\circ\lambda\circ\lambda'=p_j\circ\lambda'=\psi_j$. It follows that $\lambda\circ\lambda'\colon Y\to\Omega'$ is a map commuting with the inverse system given by the maps $p'_j$ and so $\Omega'$ is factor over $A$. We need uniqueness in order for $\Omega'$ to be universal over $A$.

In order to get uniqueness we must consider the unique map $\lambda''\colon \Omega'\to \Omega$ given by the various maps defined above. You'll end up getting $\lambda\circ\lambda'\circ\lambda''=\lambda$ by uniqueness of $\lambda$ and also if $\sigma\colon Y\to\Omega'$ is another map making the diagram commute, then $\sigma\circ\lambda''=\lambda$ and so we have $\lambda\circ\lambda' \circ \lambda''= \sigma\circ \lambda''\implies \lambda\circ\lambda' \circ \lambda''\circ \lambda'= \sigma\circ \lambda''\circ\lambda'$. Finally, it can be shown that $\lambda''\circ\lambda'=\mbox{Id}_{\Omega}$ by yet another use of the universal property and so $\sigma=\lambda\circ\lambda'$ hence $\lambda\circ\lambda'$ is unique. It then follows that $\Omega'$ is isomorphic to $\Omega$ as required.