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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4$\begingroup$ Let $f$ be the inclusion of a dense subspace and let $A = B = X$...? $\endgroup$– Qiaochu YuanJul 2, 2013 at 6:05
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$\begingroup$ @QiaochuYuan, I am a bit slow and do not understand your comment... Could you elaborate a bit? $\endgroup$– András BátkaiJul 2, 2013 at 6:26
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2$\begingroup$ @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. $\endgroup$– Qiaochu YuanJul 2, 2013 at 6:27
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$\begingroup$ @QiaochuYuan, thanks much for the obvious counterexample. You're right: I must be missing some condition. $\endgroup$– Tom LaGattaJul 2, 2013 at 6:37
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1$\begingroup$ Voting to close since this question needs further thought before continuing $\endgroup$– Yemon ChoiJul 2, 2013 at 7:02
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1 Answer
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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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3$\begingroup$ Please. This is too basic to belong here. $\endgroup$– Todd Trimble ♦Jul 2, 2013 at 7:44
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5$\begingroup$ @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. $\endgroup$ Jul 2, 2013 at 7:50
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1$\begingroup$ @AndrasBatkai: I suppose we can agree to disagree about what is "standard". $\endgroup$– Todd Trimble ♦Jul 2, 2013 at 13:36
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1$\begingroup$ @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. $\endgroup$ Jul 3, 2013 at 8:06