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To me, $\mathrm{Mat}_{m \times n}$ is isomorphic to $\mathbb{R}^{mn}$, hence is boundaryless. But this disqualified the use of Sard's theorem in this question: An exercise on Regular Value Theorem. But it seems using Sard's theorem is on the right track.

Thank you~

Community
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1LiterTears
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2 Answers2

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$\mathrm{Mat}_{m \times n}$ is no boundary for exactly the reason you stated: it is diffeomorphic to $\Bbb R^{mn}$.

With regards to your other question, you can still apply the Transversality Theorem on page 68 of Guillemin and Pollack. This is because a manifold $M$ in the usual sense is also a manifold with boundary; it just has empty boundary: $\partial M = \varnothing$. Just look at the definition of manifold with boundary and you'll see that there is no requirement that the manifold actually has boundary points, but instead the definition merely says that every point is either an interior point or a boundary point.

Henry T. Horton
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  • This this question becomes even more bizarre - since both $\mathbb{R}^n$ and $\mathrm{Mat}_{m \times n}$ are manifolds with empty boundary, how could it meet the condition that only $X$ has boundary in the Transversality Theorem on GP68? Thank you! – 1LiterTears Jun 14 '13 at 02:22
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    The second paragraph of my answer explains this. You should read the conditions in GP68 as "only $X$ is allowed to have boundary, all other manifolds in question must be boundaryless." So $X$ does not have to have a nonempty boundary. This is really a condition imposed on $Y$, $Z$, and $S$, since we are saying they cannot have boundary. $X$ can have any boundary possible, including the empty boundary! – Henry T. Horton Jun 14 '13 at 02:26
  • Thank you Henry - that is really helpful! Then, how can I show $F$ transverse {0}...? Sorry I am still stuck one step ahead... – 1LiterTears Jun 14 '13 at 02:31
  • I mean, can I use the fact that $dF$ is surjective, and {0} has dimension 0? Then by the definition of transversality, Image$(df_x)$ + $T_y(Z) = T_y(Y)$. – 1LiterTears Jun 14 '13 at 02:32
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    Yes, that works. – Henry T. Horton Jun 14 '13 at 02:42
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Yes: under its usual topology, $\mathrm{Mat}_{m\times n}$ is a manifold without boundary (a.k.a., a manifold).

Zev Chonoles
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