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Let $f : X \to Y$ be a continuous map of complete topological vector spaces. Suppose that $A \subseteq X$ and $B \subseteq Y$ are proper, dense linear subspaces, and that the restriction map $f : A \to B$ is an isomorphism. Is the original map an isomorphism?

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closed as unclear what you're asking by Tom LaGatta, Yemon Choi, Daniel Moskovich, Todd Trimble, Qiaochu Yuan Jul 2 '13 at 7:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Let $f$ be the inclusion of a dense subspace and let $A = B = X$...? – Qiaochu Yuan Jul 2 '13 at 6:05
@QiaochuYuan, I am a bit slow and do not understand your comment... Could you elaborate a bit? – András Bátkai Jul 2 '13 at 6:26
@András: Suppose $Y$ is a topological vector space and $X$ a dense subspace of it. Let $f : X \to Y$ denote the inclusion. $f$ restricts to an isomorphism $f : A \to B$ where $A = B = X$, but $f$ is not itself an isomorphism if $X$ is not all of $Y$. But the question mark in my comment is because I assume Tom forgot a condition or something. – Qiaochu Yuan Jul 2 '13 at 6:27
@QiaochuYuan, thanks much for the obvious counterexample. You're right: I must be missing some condition. – Tom LaGatta Jul 2 '13 at 6:37
Voting to close since this question needs further thought before continuing – Yemon Choi Jul 2 '13 at 7:02
up vote 4 down vote accepted

Robertson & Robertson: Topological Vector Spaces. Chapter VI (Completeness), Proposition 6 and Corollary 1.

If $E$ and $F$ are separated l.c. spaces, $t:E\to F$ continuous, linear, then there is a unique continuous linear extension $\hat t: \hat E \to \hat F$. Here $\hat E$ is the completition.

If further $t$ is an isimorphism onto $t(E)$, then $\hat t$ is an isomorphism of $\hat E$ onto $\hat t(\hat E)$.

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Please. This is too basic to belong here. – Todd Trimble Jul 2 '13 at 7:44
You mean $t$, not $f$. – The User Jul 2 '13 at 7:47
@ToddTrimble: Though this is indeed too basic for the expert, it does not belong to the standard mathematical curriculum, this is why I decided to answer. – András Bátkai Jul 2 '13 at 7:50
@AndrasBatkai: I suppose we can agree to disagree about what is "standard". – Todd Trimble Jul 2 '13 at 13:36
@TaQ: The question was about whether this trivial result has an extension to topological vector spaces. I cited the most general result I know in this context, the question is closed, I do not see the point of any further discussion here. – András Bátkai Jul 3 '13 at 8:06

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