# Does the operation of completion preserve injectivity?

It seems to me I saw a counterexample somewhere, but I can't find it, can anybody help me?

Let $\varphi:X\to Y$ be a linear continuous map of locally convex spaces, and $\widetilde{\varphi}:\widetilde{X}\to\widetilde{Y}$ the corresponding linear continuous map of their completions.

If $\varphi:X\to Y$ is injective, is $\widetilde{\varphi}:\widetilde{X}\to\widetilde{Y}$ injective as well?

I think, the answer must be "no", so another question is

Under which conditions the injectivity of $\varphi:X\to Y$ implies the injectivity of $\widetilde{\varphi}:\widetilde{X}\to\widetilde{Y}$?

• I don't have any references at hand, by I believe that there is a criterion of W. Robertson ensuring the injectivity of the extension to the compltions: $X$ should have a $0$-neighbourhood basis of absolutely convex sets $U$ such that $\varphi(U)$ is closed in $Y$. Probably, this can be found in Köthe's book on topological vector spaces. – Jochen Apr 16 '16 at 12:57

## 1 Answer

Here is a way of getting lots counterexamples. Let $\widetilde{X}$ be your favorite infinite-dimensional complete space, let $X\subset \widetilde{X}$ be a dense subspace, let $v\in\widetilde{X}\setminus X$, and let $L$ be the span of $v$. Then we can take $Y=\widetilde{X}/L$, and the quotient map $\widetilde{\varphi}:\widetilde{X}\to Y$ is not injective but its restriction to $X$ is.

I don't know of any useful conditions under which the answer is yes.