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I know that we can refered to the question How to show that the diagonal of $X\times X$ is diffeomorphic to $X$?. I have the same question with an answer, and I needed that someone tell me if it is good or not (why?)

Here's the question :

The diagonal $\Delta$ in $X \times X$ is the set of points of the fom $(x,x)$. Show that $\Delta$ is diffeomorphic to X, so $\Delta$ is a manifold if $X$ is.

Solution : First, let $ϕ:X→Δ$ $ϕ:x↦(x,x)$. This is a smooth function, as it extends easily to $Φ:R^N→R^N×R^N$, $Φ(x)=(x,x)$ which is linear. Its inverse, $ϕ^{−1}(x,x)=x$ is a restriction of the (smooth) projection function.

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    "Diffeomorphic" is already a term that only applies to (smooth) manifolds, so in order to use it here you already need to know that $\Delta$ is a (smooth) manifold. – Qiaochu Yuan May 06 '15 at 18:11
  • Does the restriction of a smooth manifold is a smooth manifold? Do I need an additional condition for this hypothesis is true? –  May 06 '15 at 18:20
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    The way I would prefer to say it is the following. The diagonal map $\Delta : X \ni x \mapsto (x, x) \in X \times X$ is clearly a smooth map. The question is to what extent it exhibits $X$ as a submanifold of $X \times X$. It turns out that it exhibits $X$ as an embedded submanifold. – Qiaochu Yuan May 06 '15 at 18:24
  • I was not asking this question in reference to the current solution. I found another solution, and wanted to know if the restriction of a smooth manifold is a smooth manifold (in the broad sense of the question). Is this true or there are examples that may contradict this hypothesis. –  May 06 '15 at 18:31
  • I don't know what you mean by "restriction of a smooth manifold." – Qiaochu Yuan May 06 '15 at 18:32
  • Ok, if I have the smooth manifold $X \times X$ and I restricted this set to $ \Delta $, does $ \Delta $ will remain a smooth manifold or not? –  May 06 '15 at 18:37
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    @J.G Without requiring $X$ to be a smooth? That is, assume $X$ is a manifold and we have a smooth structure on the product manifodld $X\times X$, does that make $\Delta$ a smooth submanifold? - That would cetainly be wrong as can be shown by embedding $\mathbb R\to\mathbb R^n$ "unsmoothly" – Hagen von Eitzen May 06 '15 at 18:44
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    @J. G: being a smooth manifold is not a property of a subset of a smooth manifold. A priori it is a structure. It is certainly not true that arbitrary subsets of smooth manifolds inherit smooth structures. – Qiaochu Yuan May 06 '15 at 18:47
  • This exactly my question, but I am not able to answer this question. –  May 06 '15 at 18:48
  • Are you able to explain why more rigorously using an example or theory of a smooth manifold? –  May 06 '15 at 18:50
  • @J.G: for starters the problem is that it is already not true that a subset of a smooth manifold, equipped with the subspace topology, is a topological manifold (note that this is a property of a topological space and not a structure); take, for example, the Cantor set in $\mathbb{R}$. Even if you take a subset that is a topological manifold it's not clear how it ought to inherit a smooth structure; take, for example, a square in $\mathbb{R}^2$, which is a topological manifold homeomorphic to $S^1$, but the natural embedding is not smooth... – Qiaochu Yuan May 06 '15 at 19:28
  • Ok, but when does a subset of smooth manifold is itself a smooth manifold. What are the conditions on this set that would allow us to say that it is itself a manifold? –  May 06 '15 at 19:42
  • I don't know if my question is clear. Certain subsets of a smooth manifold are themselves smooth manifolds (submanifold). For example, we already know that $ S^1 \subset \mathbb {R}^2 $ is a smooth manifold of $1-dimension$, and $ H= {(x, y) \in \mathbb{R^2} : x^2 + y^2 = 1~and~y>0 }$ is itself a smooth manifold (submanifold) of $1-dimension$; although $ H \subset S^1$ . The question is when does a subset of smooth manifold is itself a smooth manifold. Why are some subsets are smooth manifold, while the other does not. –  May 06 '15 at 19:51
  • I found the answer at my question. ''Open subsets of a smooth manifold are smooth manifolds. ''

    reference : http://www-personal.umich.edu/~wangzuoq/635W12/Notes/Lec%2002.pdf

     –  May 06 '15 at 20:08
  • The map $\Phi$ is NOT linear! It doesn't commute with sums. – Alex M. May 06 '15 at 20:30
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    Yes, it is $\longrightarrow$ Let $x,y \in \mathbb{R^N}~and~c \in \mathbb{R}$. Then $\Phi(x+y)=(x+y,x+y) = (x,x) + (y,y) = \Phi(x) + \Phi(y) $ and $\Phi(cy)=(cy,cy) = c(y,y) = c \Phi(y) $ –  May 06 '15 at 20:41

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