Inverse limit of algebras

Remarks

The inverse limit is closed under the operations of the product, so it is a subalgebra of $\prod_i A_i$; it can be empty when the signature has no constant symbols. The projections $\pi_i \colon \varprojlim A_j \to A_i$ satisfy $\varphi_{ij} \circ \pi_j = \pi_i$. For every algebra $B$ with homomorphisms $\psi_i \colon B \to A_i$ satisfying $\varphi_{ij} \circ \psi_j = \psi_i$, there is a unique homomorphism $\psi \colon B \to \varprojlim A_i$ with $\pi_i \circ \psi = \psi_i$, namely $\psi(b) = (\psi_i(b))_i$.

Example: let $p$ be a prime, let $I = \N$, let $A_n = \Z/p^n\Z$ as rings, and let $\varphi_{mn}$ be reduction modulo $p^m$ for $m \le n$. Reduction modulo $p^n$ followed by reduction modulo $p^m$ is reduction modulo $p^m$, so clause 3 holds. The inverse limit is the ring $\Z_p$ of $p$-adic integers.

Non-example: on $I = \set{1, 2, 3}$ with the usual order, take $A_i = \Z/2\Z$ as groups, $\varphi_{ii} = \varphi_{12} = \varphi_{23} = \id$ and $\varphi_{13} = 0$. Then $\varphi_{12} \circ \varphi_{23} = \id \ne \varphi_{13}$, so clause 3 fails and the data is not an inverse system.

Some authors allow any partially ordered set, or any small category, as the index; the construction is the same.