Let $f:V \rightarrow W$ be an onto linear map and $g:W \rightarrow V$ be an onto linear map. Does there exist a linear isomorphism from $V \rightarrow W$. i am stuck on the case when $V$ and $W$ are infinite-dimensional.
Please help.
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I'm pretty sure this requires the axiom of choice; perhaps one of our resident set theorists will come along and explain the details. – Zev Chonoles Mar 01 '13 at 05:17
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The answer to your question is yes. Let $\dim V=\lambda$ and $\dim W=\kappa$. Fix a basis $B$ for $f$. Since $f(B)$ spans $W$ there exists a a surjection of $B$ onto a basis for $W$. Thus, $\lambda\geqslant \kappa$. Using the exact same idea in reverse says that $\kappa\geqslant\lambda$. It then follows from the Schroeder-Bernstein theorem that $\lambda=\kappa$. Thus, $V$ and $W$ have the same dimension and so are isomorphic.
Alex Youcis
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Cantor-Schroeder-Bernstein is for injections, not surjections. See this math.SE thread. – Zev Chonoles Mar 01 '13 at 05:19
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@ZevChonoles The axiom of choice implies that if there is a surjection $X\to Y$ there is an injection $Y\to X$. – Alex Youcis Mar 01 '13 at 05:19
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Agreed, but my point was only that you are using the axiom of choice here, whereas the theorem itself does not. – Zev Chonoles Mar 01 '13 at 05:20
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@GerryMyerson True, true. This is why I don't think about the AC--I undoubtedly used it at least three times typing this response. – Alex Youcis Mar 01 '13 at 05:31