Let \(K\) be a CW-complex and \(X\) a space. Then the statement \(X \tau K\) (or \(K \in AE (X))\) means that for every closed subset \(A\) of \(X\) and map \(f : A \to K\), there exists a map \(F : X \to K\) which is an extension of \(f\). The main theorem of this paper states that if \(K\) is countable, \(X\) is a separable metrizable space, and \(X \tau K\), then there exists a completely metrizable separable space \(Y\) with \(X\) embedded in \(Y\) and \(Y \tau K\). In dimension theory, \(X \tau S^n\) means that \(\dim X \leq n\). For cohomological dimension over an abelian group \(G\), \(X \tau K (G,n)\) means \(\dim_G X \leq n\); here \(K(G,n)\) is an Eilenberg-MacLane complex for the group \(G\). For dimension theory [R. Engelking, Theory of dimension finite and infinite, Heldermann Verlag, Lemgo, Germany, 1995], it is a classical theorem that every metric space (separable or not) has a completion having the same dimension. This same result was proved for \(\dim_\mathbb{Z}\) by the reviewer and P. J. Schapiro [Topology Appl. 22, 221-244 (1987; Zbl 0646.54038)]. The main theorem of the current paper, which restricts only to separable metrizable spaces unlike the preceding two results, implies that if \(G\) is a countable abelian group and \(X\) is a separable metric space with \(\dim_GX \leq n\), then \(X\) can be embedded in a completely metrizable space \(Y\) with \(\dim_GY \leq n\). The author also proves that in case \(G\) is a torsion group, then the same completion theorem is true. For arbitrary abelian groups the author provides a completion \(Y\) with \(\dim_GY \leq \dim_G X + 1\).