Example 4.22.6. Let $\mathcal{C}$ be a category. Let $(X_ n)$ and $(Y_ n)$ be inverse systems in $\mathcal{C}$ over $\mathbf{N}$ with the usual ordering. Picture:

Let $a : (X_ n) \to (Y_ n)$ be a morphism of pro-objects of $\mathcal{C}$. What does $a$ amount to? Well, for each $n \in \mathbf{N}$ there should exist an $m(n)$ and a morphism $a_ n : X_{m(n)} \to Y_ n$. These morphisms ought to agree in the following sense: for all $n' \geq n$ there exists an $m(n', n) \geq m(n'), m(n)$ such that the diagram

commutes. After replacing $m(n)$ by $\max _{k, l \leq n}\{ m(n, k), m(k, l)\} $ we see that we obtain $\ldots \geq m(3) \geq m(2) \geq m(1)$ and a commutative diagram

Given an increasing map $m' : \mathbf{N} \to \mathbf{N}$ with $m' \geq m$ and setting $a'_ i : X_{m'(i)} \to X_{m(i)} \to Y_ i$ the pair $(m', a')$ defines the same morphism of pro-systems. Conversely, given two pairs $(m_1, a_1)$ and $(m_1, a_2)$ as above then these define the same morphism of pro-objects if and only if we can find $m' \geq m_1, m_2$ such that $a'_1 = a'_2$.

## Comments (0)

There are also: