Abstract. This paper has three purposes. The first is to present a direct example of a quotient of an injective space that is not itself injective. The second is to demonstrate that Theorem 3 in [2] is false. An s -dense subspace S of a. free space F is not free by virtue of F/ S having only values that are cofinal with <•>. Finally, we furnish a valid proof, independent of the aforementioned Theorem 3, that a subspace V of a free space is free if V satisfies Fuchs' countability condition. . One consequence of Theorem 7 (if it were true) is that every valuated vector space has projective dimension at most one. The fact that Theorem 7 in [1] is false was discovered (by all authors mentioned above) indirectly through the discovery of spaces with projective dimension greater than 1. We shall give a direct counterexample, by exhibiting a subspace W of a product P of homogeneous spaces such that P/ W is not injective.The value of an element x is denoted herein by \x\. Recall that if Wis a subspace of V the valuation on the quotient space V/ W is given by \x + W\ = sup{|x + w\:w E. W) for any x G V. For convenience of proof that our example is valid, we shall employ the notion of a separable subspace which was introduced in [3]. Definition 1. A subspace W of V \% separable if \x + W\ = sup{|x + wn\} for some countable sequence {wn} of elements in W.Definition 2. A valuated vector space W is said to be absolutely separable if it is separable in every containing space.Lemma 1 [3]. Any free space is absolutely separable.The next result is of independent interest and is essential for our main result.Lemma 2. Any separable subspace of an injective valuated vector space is absolutely separable.