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The entire content is rather drafty, but I am especially baffled with the last comment "and thus produces an image manifold that is a diffeomorphic copy of $X$ adjacent to the original." This sentence does not make sense to me, like why we suddenly discuss original, why it is this case, and why this matter?

$\quad$We must deal with the necessity of deforming $X$ in a mathematically precise manner. Attempting to define deformations of arbitrary point sets in $Y$ is hopeless, so we shift our point of view somewhat. Considering $X$ as an abstract manifold and its inclusion mapping $i:X\hookrightarrow Y$ simply as an embedding, we know how to deform $i$, namely by homotopy. Since embeddings form a stable class of mappings, any small homotopy of $i$ gives us another embedding $X\to Y$ and thus produces an image manifold that is a diffeomorphic copy of $X$ adjacent to the original.

1LiterTears
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1 Answers1

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If $f:X\to Y$ is a smooth embedding, then the image $f(X)$ is diffeomorphic to $X$.

Let $F:X\times (-\epsilon,\epsilon)\to Y$ be a smooth map, and write $f_t(x)$ for $F(x,t)$. Thus, we have a family of smooth maps $f_t:X\to Y$. Figuratively speaking, when $t,s\in (-\epsilon,\epsilon)$ are "close", the images $f_t(X)$ and $f_s(X)$ can be said to be "adjacent". Consider this gif of a homotopy (from wikipedia):

enter image description here

For times $t$ and $s$ that are close together, the image curves at time $t$ and time $s$ are "adjacent".

"Embeddings form a stable class of mappings" means that if $f_t$ is an embedding for some $t$, then for some neighborhood $S\subseteq(-\epsilon,\epsilon)$ of $t$, we will have that $f_s$ is an embedding for all $s\in S$.

Thus, if $X\subseteq Y$ is an embedded submanifold of $Y$ (so that $i:X\hookrightarrow Y$ is a smooth embedding), then a homotopy of $i$ (that is, a map $F:X\times(-\epsilon,\epsilon)\to Y$ with $f_0=i$) will, for $t$ sufficiently close to $0$, produce embeddings $f_t:X\to Y$, whose image manifolds $f_t(X)$ are "adjacent" to $X$ and diffeomorphic to $X$.

Zev Chonoles
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  • Haha, glad to help :) – Zev Chonoles Jul 17 '13 at 01:36
  • Hey Zev, I have been wondering your answer for a few hours.. To start with, if $f$ is a smooth embedding, we know $df_x$ is injective at all $x$, $f$ is injective hence has inverse. But how can we know $f^{-1}$ is smooth, and $f$ is surjective..? – 1LiterTears Jul 17 '13 at 05:36
  • Functions don't have inverses unless they're bijective; the statement "if $f:X\to Y$ is an embedding, then $f(X)$ is a submanifold of $Y$ that is diffeomorphic to $X$" doesn't require $f$ to be injective. However, if you restrict the codomain $Y$, obtaining a function $f:X\to f(X)$, this function *is* bijective and indeed is a diffeomorphism. Concisely, one says that $f$ (an embedding) is a diffeomorphism onto its image. See the relevant wikipedia page. – Zev Chonoles Jul 17 '13 at 06:38
  • Actually embedding means homeomorphic to the image set, so you don't have to worry about the surjectivity. – lee Jul 17 '13 at 06:50
  • Hey @lee! Yes I got that I don't need to worry about surjectivity on the image set. But how about the smoothness of $f^{-1}$? – 1LiterTears Jul 17 '13 at 06:52
  • Hi Zev, according to your wiki reference, I traced back to Warner, F.W. (1983), but it only says $f^{-1}$ is $\mathcal{C}^\infty$ by definition of diffeomorphim, but didn't prove the the smoothness of $f^{-1}$ is assumed for this case. – 1LiterTears Jul 17 '13 at 07:27
  • Dear @Jellyfish , if it's embedding, then it is locally diffeomorphism, hence the inverse is smooth. Did you know the inverse function theorem or implicit function theorem? – lee Jul 17 '13 at 11:36
  • Oh, yeah, I got it... Thanks @lee :D – 1LiterTears Jul 17 '13 at 21:58