Timeline for Is there a Whitney Embedding Theorem for non-smooth manifolds?
Current License: CC BY-SA 2.5
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| Aug 4, 2023 at 14:53 | comment | added | murray | Why is $f_i$ continuous? I presume your definition of $f_i$ is $h \circ g_i$ where: $g_i : M \rightarrow \mathbb{R}^n \cup \{\infty\}$, the latter codomain being the one-point compactification of $\mathbb{R}^n$; $g_i = f_i$ on $U_i$ and $g_i = \infty$ on $M \setminus U_i$; and $h: \mathbb{R}^n \cup \{\infty\} \rightarrow S^n$ is the usual homeomorphism. So my question is why is $g_i$ continuous? It seems to me some modfication is needed to ensure that, possibly involving a "shrinking" of the original cover. | |
| Aug 5, 2010 at 17:28 | history | answered | Andrea Ferretti | CC BY-SA 2.5 |