Let $(R_i,f_{ij})_{i \in I}$ be an inverse system of rings and ring homomorphisms. Let $(L,p_i)_{i \in I}$ be the inverse limit of $(R_i,f_{ij})_{i \in I}$ and let $(K,f_i)_{i \in I}$ be a cone into $(R_i,f_{ij})_{i \in I}$. My question is: can always find an embedding $u:K \rightarrow L$?
Let $R$ be the canonical inverse limit of $(R_i,f_{ij})_{i \in I}$. We define the map $g:K \rightarrow R$ by \begin{equation} g(k)=\displaystyle \prod_{i \in I}f_i(k). \end{equation} Since $(K,f_i)_{i \in I}$ is a cone this is clearly well-defined but I don't seem to be able to show that $g$ is injective (is it?). If it is then we can set $u=hg$ where $h$ is the (unique) isomorphism between $L$ and $R$. Then $u$ is a composition of injective morphisms and so it is injective itself. Does that work or is $g$ not always injective?
The map $u:K \to L$ need not be an embedding. For instance, consider the category $I = \emptyset$ so that your inverse system of rings is the empty system. Then the limit of this diagram is the zero ring, i.e., $L = 0$. Any ring $K$ is also a cone to the system (for silly reasons) and the unique map $u:K \to 0$ is the unique cone morphism. For any ring $K$ with $\lvert K \rvert \geq 2$ we have that $u$ is a non-injective map so you can't quite form an embedding $K \to 0$. In fact, this map $u$ has an inverse if and only if $\lvert K \rvert = 1$.