The statement can be corrected by adding one word:

> Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each **cofinal** order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is cofinal in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$  is either cofinal (and uniformly continuous) or discrete (and wildly discontinuous). 

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is cofinal unless $K$ is already complete.

*Remark 1*: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy but such a $K$ is also trivially complete. Therefore the theorem with "cofinal" and "sequentially Cauchy" are precisely equivalent.

*Remark 2*: Cohn's completion is unusual because he only considers Cauchy *sequences*. However, it does coincide with the usual topological completion when the ordered field is metrizable. An ordered field is metrizable precisely if it has countable cofinality. When a field has uncountable cofinality, all Cauchy sequences are eventually constant and Cohn's completion trivializes.