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Here is the question I am trying to tackle:

Show that the map $$f : \mathbb R P^n \to \mathbb R P^{n + 1},$$ defined by $$[p] = [p_0, \dots, p_n] \mapsto [p,0] = [p_0, \dots, p_n, 0]$$

is an embedding.

I am having a hard time in proving that it is a submanifold, what exactly should we do to prove that it is a submanifold? I think I do not understand the definition very well.

Here is the definition of a submanifold I am using:

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Here is the definition of an embedding:

enter image description here

The hint I have for the solution of this problem:

enter image description here

Could someone help me please?

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    What is your definition of embedding? The one I have in mind is an immersion which is a homeomorphism onto its image (equipped with the subspace topology). Therefore, if you wish to show that this map is an embedding, this might help https://math.stackexchange.com/questions/1952990/show-that-injective-immersion-of-a-compact-manifold-is-an-embedding. And then one can show that the image of an embedding is a submanifold, but I'm not sure if that's what you want. – Amadeus Maldonado May 01 '22 at 20:23
  • @AmadeusMaldonado I will edit my question with the definition, thanks for pointing this out. –  May 01 '22 at 22:05
  • Use that $f$ is the passage to quotients of an embedding $\mathbb{S}^n \rightarrow \mathbb{S}^{n+1}$, and that the quotients are local diffeomorphisms. – Laz May 01 '22 at 23:17
  • I do not know how to do this, you can please show me how?@Laz –  May 02 '22 at 05:59
  • @Laz how do you know that the quotients are local diffeomorphism? ..... it is clear for me that the map is 1-1 and onto but it is not clear for me how the inverse is differentiable .... is that what you are referring to? .... what about the proof that it is a submanifold? I think this is hard proof. –  May 04 '22 at 16:58
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    I am currently working on an answer. – Laz May 04 '22 at 16:59
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    @Laz I will also edit my post with the hint at the back of the book. –  May 04 '22 at 17:08
  • @Laz also here is the atlas of $RP^n$ I have https://math.stackexchange.com/questions/4384016/understanding-the-atlas-of-mathbb-rpn –  May 04 '22 at 18:19
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    I have posted an answer, if you need more details, just let me know. – Laz May 04 '22 at 18:52

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For each $n\in\mathbb{N}$ and $i=0,\dots, n$ let $U_i^n\subset \mathbb{R}P^n$ be the open subset $U_i^n = \{[p_0,\dots,p_n], p_i\neq 0\}$, and $\phi_i^n: U_i^n \rightarrow \mathbb{R}^n$ be given by $\phi_i^n([p_0,\dots,p_n]) = (\frac{p_0}{p_i}, \dots, \frac{p_{i-1}}{p_i}, \frac{p_{i+1}}{p_i},\dots,\frac{p_n}{p_i})$. Then $\mathcal{A}^n=\{(U_i^n,\phi_i^n)\}_{i=0}^n$ is a smooth atlas on $\mathbb{R}P^n$ for each $n\in\mathbb{N}$. Let us set $\phi_i^n = (x_i^1,\dots,x_i^n)$
First, notice that $f([p_0, \dots, p_n]) = f([q_0, \dots, q_n]) \implies(q_0, \dots, q_n,0)=\lambda (p_0, \dots, p_n,0), \lambda\in\mathbb{R}$, and thus $(q_0, \dots, q_n)=\lambda (p_0, \dots, p_n) \implies [p_0, \dots, p_n] = [q_0, \dots, q_n]$. So that $f$ is one-one.
To prove that $f(\mathbb{R}P^n)\subset \mathbb{R}P^{n+1}$ is a submanifold, we just use the first $n+1$ charts of $\mathcal{A}^{n+1}$. Indeed, for any $(p_0,\dots, p_n)\in\mathbb{R}^{n+1}\setminus\{0\}$, there is an $i< n+1$ for which $p_i\neq 0$. Then $U_i^{n+1}$ is an open neighborhood of $f([p_0, \dots, p_n])$. Moreover, the last coordinate of $\phi_i^{n+1}([r_0,\dots,r_{n+1}])$ vanishes if and only if $r_{n+1} = 0$. Thus, $U_i^{n+1}\cap f(\mathbb{R}P^n) = \{[p_0,\dots,p_n,0], p_i\neq 0\} = \{r\in U_i^{n+1}, x_i^{n+1}(r)=0\}$, and $f(\mathbb{R}P^n)$ is an $n$-dimensional submanifold of $\mathbb{R}P^{n+1}$.
These charts are what are called adapted charts for the submanifold $f(\mathbb{R}P^n)$, and we will use them to prove the smoothness of both $f$ and $f^{-1}$.
Notice that smoothness is a local condition, so it will be enough to prove the smoothness of any map when restricted to open neighborhoods. Take any point $p=[p_0, \dots, p_n]\in \mathbb{R}P^n$ and choose $i < n+1$ with $p_i\neq 0$. Then $(U_i^n,\phi_i^n)$ is chart in $\mathbb{R}P^n$ around $p$ and $(V,\psi)$ is a chart in $\mathbb{R}P^{n+1}$ around $f(p)$, where $V =U_i^{n+1}\cap f(\mathbb{R}P^n)$, $\psi =\pi\circ\phi_i^{n+1}|_{f(\mathbb{R}P^n)}$, $\pi$ being the projection onto the first $n$ coordinates. Next, we just compute the local expression of $f$ relative to these charts:

$ \begin{align} \psi\circ f\circ(\phi_i^n)^{-1}(x^1,\dots,x^n) &= \psi\circ f ([x^1,\dots,x^{i-1},1,x^i,\dots, x^n]) \\ &= \psi([x^1,\dots,x^{i-1},1,x^i,\dots, x^n,0])\\ &= (x^1,\dots, x^n). \end{align} $

From this, not only do we get the smoothness of $f$, but also that of $f^{-1}$, since both are locally the identity map $\mathbb{1}_{\phi_i^n(U_i^n)}$. So, $f$ is an embedding.

Laz
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    Thank you! I will read it thoroughly in a couple of hours and let you know. –  May 04 '22 at 21:58
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    Notice that I have edited my answer, to include a proof that uses the hint you were given, and your definition of embedding. The previous answer was elegant, but it required more details. I think this one is more straightforward. Anyway, if you'd like to go through the other proof, I will add it. – Laz May 05 '22 at 02:38
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    I am good with the proof using the hint and my definition of embedding. I will read it thoroughly now and let you know if I am struggling with something. –  May 05 '22 at 03:56
  • I do not understand in the hint, what it means by the "ups, over and acrosses" ..... could you please explain this to me? –  May 05 '22 at 04:01
  • For the sake of context, what book are you using? – Laz May 05 '22 at 04:02
  • A short Course in Differential Topology by Bjorn Ian Dundas –  May 05 '22 at 04:07
  • why the inverse of $f$ is smooth? I need this to complete the proof of diffeomorphism –  May 05 '22 at 04:32
  • In the hint, what does it mean saying that "by using all but the last of the standard charts on $\mathbb RP^{n+1}$"? .... what are the last standard charts on $\mathbb RP^{n+1}$? –  May 05 '22 at 04:45
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    When the author says "ups, over and acrosses", they are referring to what I call the local expression of a map between manifolds. In this case, since $f$ is such a natural map, the local expression looks like the identity map, making it look like f is a map from $\mathbb{R}P^n$ to itself. That is my best interpretation. But if you want mi honest opinion, is an unnecessary comment that is not technically 100% true, it's just making a comment on something you already understand. Not productive. – Laz May 05 '22 at 04:57
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    I will make an edit to my answer that will finally make $f$ into being locally the identity map. It should be that, but I left the chart $\psi$ have codomain the hyperplane $x^{n+1}=0$ in $R^{n+1}$ which is not exactly $R^n$, but a copy of it. All I'm saying is that a chart of an n-manifold should have n coordinates. With this in mind, to prove the smoothness of $f^{-1}$ you just need to prove that $\phi^n_i \circ f^{-1} \circ (\psi)^{-1}$ is smooth, but that is just the inverse of the identity map. – Laz May 05 '22 at 05:03
  • What the author means by "by using all but the last of the standard charts on $\mathbb{R}P^{n+1}$" is that the adapted atlas we put on $f(\mathbb{R}P^n)$ is the one resulting from the restrictions of the first $n+1$ charts of $\mathcal{A}^{n+1}$, which has $n+2$ charts. – Laz May 05 '22 at 05:12
  • So what are the standard charts in $RP^{n+1}$? –  May 05 '22 at 05:14
  • Let us continue this discussion in chat. – Laz May 05 '22 at 05:15