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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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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
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@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
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Voting to close since this question needs further thought before continuing –  Yemon Choi Jul 2 '13 at 7:02
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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

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1 Answer

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
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@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
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@AndrasBatkai: I suppose we can agree to disagree about what is "standard". –  Todd Trimble Jul 2 '13 at 13:36
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@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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