Historically, it was some time before mathematicians realised the importance of the transition maps in defining manifolds. The origins of manifolds date back to Gauss and Riemann in the 1800's. It wasn't until 1913 that Weyl gave a modern definition of surfaces, and then not until 1936 that Whitney gave the current definition of manifolds in arbitrary dimensions.
A topological manifold \(M\) is a second countable, Hausdorff topological space such that there are continuous charts \(\{\varphi_{\alpha} : U_{\alpha} \subseteq M \to V_{\alpha} \subseteq \mathbb{R}^n\}\) covering \(M\). That is,
Here \(n\) is fixed independently of the charts. The dimension of \(M\) is \(n\).
A topological manifold \(M\), is a second countable, Hausdorff space that is locally homeomorphic to Euclidean space. Briefly, for the last condition, we say \(M\) is locally Euclidean . Note that there are topological spaces that are locally Euclidean that are second countable, but not Hausdorff. There are locally Euclidean spaces that are Hausdorff, but not second countable, and there are also locally Euclidean spaces that are neither. Thus it is necessary for us to require both Hausdorff and second countable.
Let \(M\) be a sub-manifold of Euclidean space. Then the collection open sets on \(M\) are a second countable, Hausdorff topology. The inverses of the local parametrisations \(\varphi_{\alpha}\) are charts.
Write \(\psi_{\alpha} : V_{\alpha} \subseteq \mathbb{R}^k \to U_{\alpha} \subseteq M \subseteq \mathbb{R}^n\) for the local parametrisations. By definition, each \(\psi_{\alpha}\) is a homeomorphism, hence \(\varphi_{\alpha} := \psi_{\alpha}^{-1}\) is a homeomorphism. Also by definition of \(M\), the local parametrisations cover \(M\); that is, \(M = \cup_{\alpha} U_{\alpha}\). Thus conditions 1. and 2. for a topological manifold are satisfied.
If you think about it a little, we could have just as easily defined a topological manifold to be a second countable, Hausdorff topological space \(M\) such that there are continuous local parametrisations \(\{\psi_{\alpha} : V_{\alpha} \subseteq \mathbb{R}^k \to U_{\alpha} \subseteq M \subseteq \mathbb{R}^n\}\) covering M. That is,
Given local charts, the inverses are local parametrisations, and vice-versa given local parametrisations the inverses are local charts.
It was much more convenient to define sub-manifolds using local parametrisations. This is because local parametrisations are maps from open sets of Euclidean space into another Euclidean space and imposing the condition of differentiability with injective differential is straightforward. If we'd used local charts instead, we would need to invoke the implicit function theorem to get differentiability of the inverses, i.e. the local charts. This is indeed possible but it does make the definition a little harder to follow.
So whey then are manifolds defined by charts rather than local parametrisations? I have no idea! Presumably it was some historical convention that stuck. In the end it makes no difference which way we define manifolds and the current convention is to use local charts.
An atlas for a topological manifold is a collection of charts \(\{\varphi_{\alpha} : U_{\alpha} \subseteq M \to V_{\alpha} \subseteq \mathbb{R}^n\}\) covering \(M\).
Let \(M = \mathbb{S}^2 \subseteq \mathbb{R}^3\). Then \(M\) is an embedded sub-manifold, hence is a manifold. One possible atlas is given by covering \(M\) by six hemispheres \[ \{\phi_N, \phi_S, \phi_E, \phi_W, \phi_F, \phi_B\}. \]
Another atlas is to cover \(M\) by stereographic projection which requires only two charts, \[ \{\phi_+, \phi_-\}. \] There are many more atlases - eg. by using spherical polar coordinates to cover \(M\) as a surface of revolution.
The passage from topological to smooth manifolds turns out to be surprisingly subtle. The first issue is that we can't simply replace homeomorphism with diffeomorphism in the definition of topological manifold. The problem is that a local chart \(\varphi : U \to V\) is defined on an open subset \(U \subseteq M\) and \(M\) won't have any sort of linear structure in general, hence no notion of differentiability. We also can't work with the local parametrisations either: even though they are defined on open sets of Euclidean space, they take values in \(M\) and again there's no linear structure for us to develop calculus on \(M\).
For embedded sub-manifolds we got around this by taking local parametrisations to be differentiable functions into the ambient \(\mathbb{R}^n\) whose values lie in \(M\). Now that we no longer have an ambient \(\mathbb{R}^n\) we need a different approach. We've already seen the idea though! The idea is to use transition functions. The transition functions provide a compatibility between different charts and hence our definition will not be one for individual charts, but collections of charts.
Let \(\varphi_{\alpha} : U_{\alpha} \to V_{\alpha}\) and \(\varphi_{\beta} : U_{\beta} \to V_{\beta}\) be charts. The transition map from \(\beta\) to \(\alpha\) is the map \[ \tau_{\alpha\beta} = \varphi_{\alpha} \circ \varphi_{\beta}^{-1}|_{\varphi_{\beta} (U_{\alpha\beta})} : \varphi_{\alpha} (U_{\alpha\beta}) \to \varphi_{\beta} (U_{\beta\alpha}) \] where \(U_{\alpha\beta} = U_{\beta\alpha} = U_{\alpha} \cap U_{\beta}\).
Note that \(U_{\alpha}, U_{\beta}\) are open sets hence their intersection, \(U_{\alpha\beta}\) is open. Then since \(\varphi_{\alpha}\) is a homeomorphism. \[ \varphi_{\alpha}(U_{\alpha\beta}) = (\varphi_{\alpha}^{-1})^{\ast} (U_{\alpha\beta}) \] is open. Likewise, \(\varphi_{\beta}(U_{\alpha\beta})\) is open.
Moreover, since \(\varphi_{\alpha}\) and \(\varphi_{\beta}^{-1}\) are homeomorphisms, so too is \(\tau_{\alpha\beta}\). As a composition of continuous functions, \(\tau_{\alpha\beta}\) is continuous. In fact, we have \[ \tau_{\alpha\beta} \circ \tau_{\beta\alpha} = (\varphi_{\alpha} \circ \varphi_{\beta}^{-1}) \circ (\varphi_{\beta} \circ \varphi_{\alpha}^{-1}) = \operatorname{Id}. \] Thus \(\tau_{\alpha\beta}^{-1} = \tau_{\beta\alpha}\) is also continuous being the composition of continuous functions.
Note in particular that the inverse of a transition function is itself a transition function. Thus to prove transition functions are homeomorphisms we only needed to show they are all continuous, hence they all have continuous inverses.
A \(C^{\infty}\) ( smooth) atlas for a topological manifold \(M\) is an atlas \(\{\varphi_{\alpha} : U_{\alpha} \subseteq M \to V_{\alpha} \subseteq \mathbb{R}^n\}\) for \(M\) such that every transition map \(\tau_{\alpha\beta} : \varphi_{\alpha}^{\ast}(U_{\alpha\beta}) \to \varphi_{\beta}^{\ast}(U_{\beta\alpha})\) is \(C^{\infty}\).
Since \(\tau_{\alpha\beta}, \tau_{\beta\alpha}\) are \(C^{\infty}\), \(\tau_{\alpha\beta}^{-1} = \tau_{\beta\alpha}\) is \(C^{\infty}\) and just from assuming smoothness of transition functions, we automatically get that each transition function is a diffeomorphism.
Let \(M \subseteq \mathbb{R}^n\) be an embedded sub-manifold. Then \(M\) is covered by \(C^{\infty}\) charts \(\psi_{\alpha} : V_{\alpha} \to U_{\alpha}\) with injective differential. Then \(\varphi_{\alpha} = \psi_{\alpha}^{-1} : U_{\alpha} \to V_{\alpha}\) gives local charts covering \(M\). Moreover, we already proved (using the Inverse Function Theorem) that the sub-manifold transition functions \(\mu_{\alpha\beta} = \psi_{\alpha}^{-1} \circ \psi_{\beta}\) are \(C^{\infty}\). But then \[ \tau_{\alpha\beta} = \varphi_{\alpha} \circ \varphi_{\beta}^{-1} = \psi_{\alpha}^{-1} \circ \psi_{\beta} = \mu_{\alpha\beta} \] is \(C^{\infty}\).
Thus any cover of \(M\) by local parametrisations gives a \(C^{\infty}\) atlas \(\{\varphi_{\alpha} = \psi_{\alpha}^{-1} : U_{\alpha} \to V_{\alpha}\}\) for \(M\).
A particular case of an embedded sub-manifold is that of an open set \(U \subseteq \mathbb{R}^n\). There is just one chart needed to form at atlas - the inclusion map \(U \subseteq \mathbb{R}^n\).
General Linear Group
Let \(M = \text{GL}(V)\) be the set of invertible linear transformations \(V \to V\) of a finite dimensional vector space. Recall that the set \(L(V)\) of all linear transformations is a finite dimensional vector space of dimension \(n^2\) where \(n\) is the dimension of \(V\). The map \[ T \mapsto \det T \] is continuous \(L(V) \to \mathbb{R}\); since for example, \(\det T\) is a polynomial in the components of \(T\) with respect to a basis.
Now, \(T \in \text{GL}(V)\) if and only if \(\det T \neq 0\), hence \(\text{GL}(V) = \det^{\ast} (U)\) where \(U = \{x \neq 0\} \subseteq \mathbb{R}\) is open. Thus \(\text{GL}(V)\) is an open subset of \(L(V)\) hence is a manifold.
The unit circle \(\mathbb{S}^1\).
We've already seen that \(\mathbb{S}^n\) is an embedded sub-manifold in \(\mathbb{R}^{n+1}\) and hence in particular that \(\mathbb{S}^n\) is a \(C^{\infty}\) manifold. It may be instructive to work through the specific details explicitly.
For charts covering \(\mathbb{S}^1\) we take \[ \varphi_{\pm}^{-1} : \theta \in (0, 2\pi) \mapsto \pm (\cos \theta, \sin \theta) \] That is the charts are the inverses of the local parametrisations \(\varphi_{\pm}^{-1}\). We give the local parametrisations rather than the charts since it is a little awkward to write down the charts as functions of \((x, y) \in \mathbb{S}^1\) (for example this would often be expressed using the \(\arctan_2\) function).
The domain \(U_+\) of \(\varphi_+\) is the range of \(\varphi_+^{-1}\); namely \(U_+ = \mathbb{S}^1 \backslash \{(1, 0)\}\). That \(\varphi_+^{-1}\) is invertible follows since it is injective - it is of course onto it's range! Similarly, the domain \(U_-\) of \(\varphi_1\) is the range of \(\varphi_-^{-1}\); namely \(U_- = \mathbb{S}^1 \backslash \{(-1, 0)\}\) and \(\varphi_-^{-1}\) is also invertible. To see injectivity note for example that \(\cos\) is not injective itself but if \(\cos \theta_1 = \cos \theta_2\) then either \(\theta_1 = \theta_2\) or \(\sin \theta_1 \neq \sin \theta_2\).
Both of \(U_{\pm}\) are open sets, equal to \((\mathbb{R}^2 \backslash \{\pm (1, 0)\}) \cap \mathbb{S}^1\). That \(U_+ \cup U_-\) covers \(\mathbb{S}^1\) follows since \(U_+ = \mathbb{S}^1 \backslash \{(1, 0)\}\) and \((1, 0) \in U_-\).
The transition function \(\tau_{+-} (\theta) = \varphi_+ \circ \varphi_-^{-1}\) is defined for \(\theta \in (0, 2\pi)\) such that \(-(\cos\theta, \sin\theta) \neq (1, 0)\). That is for \(\theta \in (0, \pi) \cup (\pi, 2\pi)\). For \(\theta \in (0, \pi)\), the range of \(\varphi_-^{-1}\) is \[ \operatorname{range}\varphi_-^{-1}|_{(0,\pi)} = \{(x, y) \in \mathbb{S}^1 : y < 0\} \] parametrised counter-clockwise. For \((x, y)\) in this range, \[ \varphi_+(x, y) = \eta \in (\pi, 2\pi) \] where \[ \varphi_+^{-1}(\eta) = (\cos \eta, \sin \eta) = (x, y) = \varphi_-^{-1}(\theta) = (-\cos\theta, -\sin\theta) \]. Thus \(\eta = \theta + \pi\). That is, \[ \tau_{+-} (\theta) = \theta + \pi \] is \(C^{\infty}\) on \((0, \pi)\).
Similar considerations show that \(\tau_{+-}\) s \(C^{\infty}\) on \((\pi, 2\pi)\) and that \(\tau_{-+}\) is also \(C^{\infty}\).
The definition of a manifold accounts for the fact that there could be many different smooth atlases for a topological manifold. It turns out that in some cases, there are incompatible atlases giving a topological manifold multiple differentiable structures. This is a rather surprising and subtle point.
A \(C^{\infty}\) (or smooth) manifold is a topological manifold \(M\) equipped with a maximal \(C^{\infty}\) atlas .
A maximal atlas \(A\) is an atlas that cannot be enlarged by adding any more local charts to \(A\). That is, for every continuous local chart \(\varphi \notin A\), there is a chart \(\psi \in A\) such that the transition function \(\psi^{-1} \circ \varphi\) is not \(C^{\infty}\). There are some very interesting consequences of this fact, and we briefly describe the famous of these below.
To show that a topological manifold is a differentiable manifold, it's sufficient to give a \(C^{\infty}\) atlas. The maximal atlas is then just the collection of all continuous local charts compatible with the atlas. There are some details to fill in here, but we leave that as an exercise. From a practical point, this is all we need to know. It's worth reading the below discussion, but if you find struggle with it, you can skip over it for now without any great consequence.
Let \(M\) be a topological manifold and let \(\varphi_{\alpha}, \varphi_{\beta}\) be two local charts - i.e. homeomorphisms from an open subset of \(M\) to an open subset of \(\mathbb{R}^n\). We say \(\varphi_{\alpha}\) and \(\varphi_{\beta}\) are compatible if the transition map \(\tau_{\alpha\beta}\) is a diffeomorphism. That is, if \(\tau_{\alpha\beta}\) and \(\tau_{\alpha\beta}^{-1} = \tau_{\beta\alpha}\) are \(C^{\infty}\).
Thus an atlas is a differentiable atlas if and only if every pair of charts are compatible. Note also that if \(\tau_{\alpha\beta}\) is not defined (i.e. if \(U_{\alpha\beta} = \emptyset\)) then we define \(\varphi_{\alpha}\) and \(\varphi_{\beta}\) to be compatible. This is reasonable in the sense that \(U_{\alpha\beta} = \emptyset\) then there is no compatibility condition required and so it's vacuously true. Ultimately it's mostly for convenience otherwise we'd be forever making statements like "either \(\varphi_{\alpha}\) and \(\varphi_{\beta}\) are compatible or \(U_{\alpha\beta}\) is empty".
That is, if \(\varphi\) is a local chart not contained in \(A\), then \(\varphi\) is not compatible with some \(\psi \in A\). By Zorn's lemma, maximal \(C^{\infty}\) atlases exist and every \(C^{\infty}\) atlas is contained in a unique \(C^{\infty}\) atlas.
Let \(A, B\) be \(C^{\infty}\) atlases such that for every \(\varphi \in A\) and every \(\psi \in B\), \(\varphi\) and \(\psi\) are compatible. Then \(A \cup B\) is a \(C^{\infty}\) atlas.
This is more or less immediate. Given any \(\varphi, \psi \in A \cup B\) if both are in \(A\), or both are in \(B\), they are compatible since \(A, B\) are \(C^{\infty}\) atlases. If one is in \(A\) and one is \(B\) they are compatible by the hypotheses of the lemma.
The idea of maximal atlas is a somewhat subtle point - it may be hard to see how there could be more than one such atlas, but it turns out to be a very interesting phenomena.
The topological manifolds \(\mathbb{R}^4\) and \(\mathbb{S}^7\) admit multiple different maximal \(C^{\infty}\) atlases - also known as differentiable structures.
Note that one differentiable structure for \(\mathbb{R}^4\) is given by taking the maximal atlas containing the chart \(\operatorname{Id} : \mathbb{R}^4 \to \mathbb{R}^4\). This atlas consists of all diffeomorphisms \(\varphi : U \to V\) for \(U, V \subseteq \mathbb{R}^4\) open sets. This is the standard atlas .
It can be quite a shock to learn that there could be a different atlas of charts \(\{\varphi_{\alpha} : U_{\alpha} \subseteq M = \mathbb{R}^4\to V_{\alpha} \subseteq \mathbb{R}^4\}\) such that all transition maps \(\tau_{\alpha\beta}\) are diffeomorphisms! This means in particular that although \(\varphi_{\alpha} \circ \varphi_{\beta}^{-1}\) is a diffeomorphism, neither \(\varphi_{\alpha}\) not \(\varphi_{\beta}\) can be a diffeomorphism! Otherwise, they would be part of the standard differentiable structure. The non-standard differentiable structures known as exotic differentiable structures .
Similar remarks apply to \(\mathbb{S}^7\) - the standard differentiable structure is the maximal atlas obtained by the sub-manifold structure.
The proof of the theorem is beyond the scope of this course, but hopefully it pique's your interest! To further make the point, a topological space has at most one topological structure; in other words, a topological space has at most one maximal atlas. This is essentially because there is no comp ability condition between continuous charts. But the same is not true of maximal \(C^{\infty}\) atlases in general.
But the situation is quite strange since for example, no other \(\mathbb{R}^n\) besides \(\mathbb{R}^4\) admits exotic differentiable structures. In the case of spheres, there are results ruling out exotic structures in some dimensions, and results showing multiple structures in other dimensions. The particular case \(n = 4\) is somewhat unique and it is an open problem as to whether it admits exotic differentiable structures or not.
The real projective space , \(\mathbb{RP}^n\) of dimension \(n\) is the set of lines in \(\mathbb{R}^{n+1}\). That is, \[ \mathbb{RP}^n = \big\{V \subseteq \mathbb{R}^{n+1}, V \text{ vector subspace}, \> \operatorname{dim} V = 1\big\}. \]
This section is devoted to a detailed explanation of the following theorem:
The real projective space \(\mathbb{RP}^n\) is a \(C^{\infty}\) manifold of dimension \(n\).
Since \(V \in \mathbb{RP}^n\) has dimension one, any vector \(v \in V\) is a basis for \(V\). In particular, taking a unit length vector we obtain a function \[ \pi : v \in \mathbb{S}^n \mapsto \operatorname{span} \{v\} \subseteq \mathbb{RP}^n. \]
The map \(\pi\) is surjective.
Let \(V \in \mathbb{RP}^n\) and let \(v \in V\), \(v \neq 0\). Then \[ \pi\left(\frac{v}{\|v\|}\right) = \operatorname{span} \left\{\frac{v}{\|v\|}\right\} = V. \]
The antipodal map \(A : \mathbb{S}^n \to \mathbb{S}^n\) is the function \(A : x \mapsto -x\). We say that \(A(x)\) is the antipodal point to \(x\), or that the points \(x, -x\) are antipodal.
Note that \(A^2(x) = A \circ A(x) = x\) so that \(x\) is antipodal to \(A(x)\).
The map \(\pi : \mathbb{S}^n \to \mathbb{RP}^n\) is a \(2:1\) covering. That is, for every \(V \in \mathbb{RP}^n\), \(\pi^{\ast} (\{V\})\) is a pair of \(2\) antipodal points. That is \(\pi^{\ast}(\{V\}) = \{\pm x\}\) for some \(x \in \mathbb{S}^n\).
Let \(V \in \mathbb{RP}^n\). Then \(x \in \pi^{\ast} (\{V\})\) if and only if \(\operatorname{span} \{x\} = V\). Since \(V\) is one-dimensional, if \(x, y \in \pi^{\ast}(\{V\})\) with \(x \neq 0\), then \(y = c x\) for some \(c \in \mathbb{R}\). But \(x, y \in \mathbb{S}^n\) hence \[ 1 = \|y\| = \|cx\| = |c|\|x\| = |c| \] Thus \(c = \pm 1\) and \(y = \pm x\).
We will use the surjection \(\pi\) to induce a topology on \(\mathbb{RP}^n\) from the topology of \(\mathbb{S}^n\). This topology is known as the quotient topology .
A set \(U \subseteq \mathbb{RP}^n\) is open if and only if \(V = \pi^{\ast}(U)\) is an open set \(V \subseteq \mathbb{S}^n\).
The set of open subsets in \(\mathbb{RP}^n\) is a topology and \(\pi : \mathbb{S}^n \to \mathbb{RP}^n\) is continuous.
That \(\pi\) is continuous is immediate from the definition of topological continuity since \(U \subseteq \mathbb{RP}^n\) is open if and only if \(\pi^{\ast}(U)\) is open.
Show that for any sets \(X, Y\), a function \(f : X \to Y\), and any collection of sets \(\{U_{\alpha}\}\),
\begin{align*} f^{\ast}\left(\bigcup_{\alpha} U_{\alpha}\right) &= \bigcup_{\alpha} f^{\ast} (U_{\alpha}) \\ f^{\ast}\left(\bigcap_{\alpha} U_{\alpha}\right) &= \bigcap_{\alpha} f^{\ast} (U_{\alpha}) \end{align*}Thus projective space is a topological space. Next we exhibit local charts showing it is a topological manifold
For \(i = 1, \dots, n+1\), let \(\tilde{U}_i \subseteq \mathbb{S}^n\) be set \[ \tilde{U}_i = \{(x_1, \dots, x_{n+1}) \in \mathbb{S}^n : x_i \neq 0\}. \] Define \[ \tilde{\varphi}_i : x \in \tilde{U}_i \mapsto \left(\frac{x_1}{x_i}, \dots, \widehat{\frac{x_i}{x_i}}, \dots, \frac{x_{n+1}}{x_i}\right) \in \mathbb{R}^n \] where \(\widehat{\tfrac{x_i}{x_i}}\) means the \(i\)'t entry is omitted. Thus,
\begin{align*} \tilde{\varphi}_1 (x) &= \frac{1}{x_1} \left(x_2, \dots, x_{n+1}\right) \\ \tilde{\varphi}_2 (x) &= \frac{1}{x_2} \left(x_1, x_3, \dots, x_{n+1}\right) \\ \vdots \quad &= \quad \vdots \\ \tilde{\varphi}_n (x) &= \frac{1}{x_n} \left(x_1, \dots, x_{n-1}, x_{n+1}\right) \\ \tilde{\varphi}_{n+1} (x) &= \frac{1}{x_{n+1}} \left(x_1, \dots, x_n\right) \\ \end{align*}More succinctly, given \(x \in \mathbb{R}^{n+1}\), let \(\widehat{x_i} \in \mathbb{R}^n\) denote the point \[ \widehat{x_i} = (x_1, \dots, x_{i-1}, x_i, \dots, x_n). \] Then \[ \tilde{\varphi}_i (x) = \frac{\widehat{x_i}}{x_i}. \]
The affine charts for \(\mathbb{RP}^n\) are the functions, \[ \varphi_i : V \in U_i \mapsto \tilde{\varphi}_i(x) = \left(\frac{x_1}{x_i}, \dots, \widehat{\frac{x_i}{x_i}}, \dots, \frac{x_{n+1}}{x_i}\right) \in \mathbb{R}^n \] where \(U_i = \pi(\tilde{U}_i)\) and \(x = (x_1, \dots, x_{n+1}) \in \pi^{\ast} (\{V\}) \subseteq \mathbb{S}^n\).
The affine charts are well defined independently of the choice of \(x \in \pi^{\ast}(\{V\})\).
We have \(\pi^{\ast}(\{V\}) = \{\pm x\}\) and \[ \tilde{\varphi}_i (-x) = \frac{\widehat{-x_i}}{-x_i} = \frac{\widehat{x_i}}{x_i} = \tilde{\varphi}_i (x). \]
The affine charts give \(\mathbb{RP}^n\) the structure of a topological manifold. That is, they are homeomorphisms \(U_i \to \mathbb{R}^n\) and the \(U_i\) forms an open cover \(\mathbb{RP}^n\).
For ease of reading, let's break the theorem into two separate lemmas.
The sets \(\{U_i\}_{i=1}^{n+1}\) are an open cover for \(\mathbb{RP}^n\).
First let us see that \(U_i\) is open. The sets \(\tilde{U}_i \subseteq \mathbb{S}^n\) are open, hence by definition \[ \pi^{\ast}(U_i) = \pi^{\ast}(\pi(\tilde{U}_i)) = \tilde{U}_i \] is open. Note that in general, \(f^{\ast}(f(U)) \neq U\) however, in this case it is true since \(\tilde{U}_i\) is closed under the antipodal map; i.e. \(A(\tilde{U}_i) = \tilde{U}_i\).
The sets \(\{U_1, \dots U_{n+1}\}\) cover \(\mathbb{RP}^n\) since the sets \(\{\tilde{U}_1, \dots, \tilde{U}_{n+1}\}\) cover \(\mathbb{S}^n\), and \(\tilde{U}_i = \pi(U_i)\) with \(\pi\) is surjective. To see why \(\{\tilde{U}_1, \dots, \tilde{U}_{n+1}\}\) covers \(\mathbb{S}^n\), just observe that if \(x \in \mathbb{S}^n\), then \(x \neq 0\) hence \(x_i \neq 0\) for some \(i\), hence \(x \in \tilde{U}_i\) for some \(i\).
Show that the surjection \(\pi\) satisfies \[ \pi^{\ast}(\pi(S)) = S \] for any subset \(S \subseteq \mathbb{S}^n\) closed under the antipodal map; i.e. \(A(S) \subseteq S\). Hint : use the fact that \(\pi^{\ast}(V) = \{\pm x\}\) for any \(V \in \mathbb{RP}^n\).
Show that the surjection \(\pi : \mathbb{S}^n \to \mathbb{RP}^n\) is an open map . That is \(\pi(U) \subseteq \mathbb{RP}^n\) is open for every open set \(U \subseteq \mathbb{S}^n\).
The charts \(\varphi_i\) are homeomorphisms.
First let us show that each \(\varphi_i\) is invertible. Define \[ \psi_i : p \in \mathbb{R}^n \mapsto \frac{1}{\sqrt{1 + \|p\|^2}} \big(p_1, \dots, p_{i-1}, 1, p_{i+1}, \dots, p_{i+n}\big) \in \mathbb{S}^n. \] Then \(\varphi_i^{-1} = \pi \circ \psi_i\): since recalling that \(\varphi_i(V) = \tilde{\varphi}_i(x)\) where \(x \in \pi^{\ast}\{V\}\) we have \(\varphi \circ \pi = \tilde{\varphi}_i\) and hence
\begin{align*} \varphi_i \circ (\pi \circ \psi_i) (p) &= \tilde{\varphi}_i \circ \psi_i (p) \\ &= \tilde{\varphi}_i\left(\frac{p_1}{\sqrt{1 + \|p\|^2}}, \dots, \frac{p_{i-1}}{\sqrt{1 + \|p\|^2}}, \frac{1}{\sqrt{1 + \|p\|^2}}, \frac{p_{i+1}}{\sqrt{1 + \|p\|^2}}, \dots, \frac{p_{n+1}}{\sqrt{1 + \|p\|^2}}\right) \\ &= \tilde{\varphi}_i\left(\sqrt{1+\|p\|}^2\frac{p_1}{\sqrt{1 + \|p\|^2}}, \dots, \sqrt{1+\|p\|}^2\frac{p_{n+1}}{\sqrt{1 + \|p\|^2}}\right) \\ &= (p_1, \dots, p_n) = p. \end{align*}Here, recall that the \(k\)-th component, \(k = 1, \dots, n\) of \(\varphi_i\) is \(x_k/x_i\) and \(\varphi_i\) omits the \(i\)'th component of it's argument. Conversely, letting \(x \in \pi^{\ast} (\{V\})\) for \(V \in U_i\) we have \(x \in \tilde{U}_i\) so that \(x_i \neq 0\) for some \(i\). For convenience, by replacing \(x \in V\) by \(-x \in V\) if necessary, we may assume that \(x_i > 0\). Then
\begin{align*} (\pi \circ \psi_i) \circ \varphi_i (V) &= (\pi \circ \psi_i) \circ \tilde{\varphi}_i (x_1, \dots, x_{n+1}) \\ &= (\pi \circ \psi_i) \left(\frac{x_1}{x_i}, \dots, \frac{x_{i-1}}{x_i}, \widehat{\frac{x_i}{x_i}}, \frac{x_{i+1}}{x_i}, \dots, \frac{x_{n+1}}{x_i}\right) \\ &= \pi\left(\frac{1}{\sqrt{1+\|p\|^2}}\left(\frac{x_1}{x_i}, \dots, \frac{x_{i-1}}{x_i}, 1, \frac{x_{i+1}}{x_i}, \dots, \frac{x_{n+1}}{x_i}\right)\right) \end{align*}where \[ p = \frac{1}{x_i}\left(x_1, \dots, x_{i-1}, x_{i+1}, \dots, x_{n+1}\right). \]
Since \(x \in \mathbb{S}^n\), \(\sum_{k=1}^{n+1} x_k^2 = 1\) and hence \(p\) satisfies
\begin{align*} 1 + \|p\|^2 &= 1 + \frac{1}{x_i^2} \sum_{j\neq i} x_j^2 \\ &= 1 + \frac{1}{x_i^2} (1 - x_i^2) \\ &= \frac{1}{x_i^2}. \end{align*}Therefore,
\begin{align*} (\pi \circ \psi_i) \circ \varphi_i (V) &= \pi\left(\frac{1}{\sqrt{1+\|p\|^2}}\left(\frac{x_1}{x_i}, \dots, \frac{x_{i-1}}{x_i}, 1, \frac{x_{i+1}}{x_i}, \dots, \frac{x_{n+1}}{x_i}\right)\right) \\ &= \pi\left(\frac{1}{\sqrt{1/x_i^2}}\left(\frac{x_1}{x_i}, \dots, \frac{x_{i-1}}{x_i}, 1, \frac{x_{i+1}}{x_i}, \dots, \frac{x_{n+1}}{x_i}\right)\right) \\ &= \pi (x_1, \dots, x_{n+1}) \\ &= V \end{align*}since \(x_i > 0\) implies \(\sqrt{x_i^2} = x_i\). If instead, we had \(x_i < 0\), then \((\pi \circ \psi_i) \circ \varphi_i(V) = \pi(-x) = V\) giving the same result.
Thus \(\varphi_i\) is a bijection \(U_i \to \mathbb{R}^n\). To finish we need to show that \(\varphi_i\) is continuous and that \(\varphi_i^{-1} = \pi \circ \psi_i\) is continuous. The latter is immediate since \(\pi\) is continuous and \(\psi_i\) is continuous as a map into \(\mathbb{R}^{n+1}\) (with range lying in \(\mathbb{S}^n\)), hence \(\psi_i\) is continuous as a map into \(\mathbb{S}^n\) with the induced topology.
For the former, \(\tilde{\varphi}_i\) is the restriction to \(\mathbb{S}^n\) of a continuous defined on \(\mathbb{R}^{n+1} \backslash \{x_i = 0\}\) hence is continuous. That \(\varphi_i\) is continuous follows from the definition of the topology on \(\mathbb{RP}^n\).
Show that a function \(f : \mathbb{RP}^n \to \mathbb{R}^k\) is continuous if and only if \(f \circ \pi : \mathbb{S}^n \to \mathbb{R}^k\) is continuous. Hint : use the topological definition of continuity and the previous exercise that \(\pi\) is an open map.
Lastly to show that projective space is a \(C^{\infty}\) manifold, we need to show that the transition functions are \(C^{\infty}\).
The transition functions \(\tau_{ij} = \varphi_i \circ \varphi_j^{-1}\) are \(C^{\infty}\) on their domains.
It suffices to show for example that \(\tau_{12}\) is \(C^{\infty}\). The other cases are similar. Note also that \(\tau_{ii} = \operatorname{Id}\) so the cases \(i=j\) are immediate.
Recall the transition maps are
\begin{align*} \tau_{12} (p) &= \varphi_1 \circ \varphi_2^{-1} (p) = \varphi_1 \circ \pi \circ \psi_2 (p) \\ &= \tilde{\varphi}_1 \circ \psi_2(p_1, \dots, p_n) \end{align*}Recall also that the domain, \(\varphi_2 (U_{12})\) of \(\tau_{12}\) is the set of \(p \in \mathbb{R}^n\) such that \(\varphi_2^{-1}(p) = \pi \circ \psi_2(p) \in U_1\) where \(U_1 = \pi(\tilde{U}_1) = \pi(\{x_1 \neq 0\})\). Since \(\psi_2(p_1, \dots, p_n) = \tfrac{1}{\sqrt{1+\|p\|^2}} (p_1, 1, p_2, \dots, p_n)\), the domain is the set \(\varphi_2(U_{12}) = \{p_1 \neq 0\}\).
Now directly from the definition of \(\varphi_1, \psi_2\) we compute, for \(p\) with \(p_1 \neq 0\):
\begin{align*} \tau_{12}(p) &= \tilde{\varphi}_1 \circ \psi_2(p_1, \dots, p_n) \\ &= \tilde{\varphi}_1\left(\frac{p_1}{\sqrt{1+\|p\|^2}}, \frac{1}{\sqrt{1+\|p\|^2}}, \frac{p_2}{\sqrt{1+\|p\|^2}}, \dots, \frac{p_n}{\sqrt{1+\|p\|^2}}\right) \\ &= \frac{\sqrt{1+\|p\|^2}}{p_1} \left(\frac{1}{\sqrt{1+\|p\|^2}}, \frac{p_2}{\sqrt{1+\|p\|^2}}, \dots, \frac{p_n}{\sqrt{1+\|p\|^2}}\right) \\ &= \left(\frac{1}{p_1}, \frac{p_2}{p_1}, \dots, \frac{p_n}{p_1}\right) \end{align*}which is indeed \(C^{\infty}\) on it's domain \(\{p_1 \neq 0\}\).