Here we deal with the the "avoid feverish tomato map"

Problem 1: Find smooth bijection \(f: \mathbb{R} \rightarrow \mathbb{R}\) whose inverse is not smooth. Then add another condition to make \(f^{-1}\) smooth and prove it.

Solution: Let \(f(x)=x^3\). It has inverse \(g(x)=x^{\frac{1}{3}}\). This does not have continuous derivative at \(x=0\). So it is not smooth. If we add the condition that \(f\) has nonzero derivative, then we can conclude that its inverse is smooth.

Problem 2: Find a diffeomorphism between \(S^1 \subset \mathbb{R}^2\) and \(\mathbb{RP}^1\).

Solution: Consider charts \(U_1, \varphi_1\) and \(U_2, \varphi_2\) on \(S^1 \setminus \{(0,1)\}\) and \(S^1 \setminus \{(0,-1)\}\) respectively. Where \(\varphi_1(x,y) = \frac{x}{1-y}\) and \(\varphi_2(x,y) = \frac{x}{1+y}\). Similarly, we have charts \(V_1, \psi_1\)and \(V_2, \psi_2\) on \(\mathbb{RP}^1\) where \(V_1 =\{a\neq0\}\) and \(V_2 = \{b\neq0\}\) with \(\psi_1(ax+by=0) = \frac{b}{a}\) and \(\psi_2(ax+by=0) = \frac{a}{b}\). The diffeomorphism between our manifolds is as follows: \(F:S^1 \rightarrow \mathbb{RP}^!\) where \(F(a,b) = ((1-b)x + ay=0)\) whenever \(b \neq 1\) and \(F(a,b) = (y = 0)\) when \(b =1\). To check this is a diffeomorphism. We need to verify that \(\psi \circ F \circ \varphi^{-1}\) is smooth for any of our chart functions.

Given \(s = \frac{a}{1-b} \in \mathbb{R}\), observe that \(varphi_1^{-1}(s) = (a,b) \in S^1\) and \(F(a,b) = ((1-b)x + ay=0)\) and \(\psi_1((1-b)x+ay)=\frac{a}{1-b}\), this is smooth since \(b\neq 0\). If we replaced \(\psi_1\) by \(\psi_2\) we get \(\psi_2((1-b)x+ay=0) = \frac{1-b}{a}\). This is smooth since \(a\neq0\) in \(U_1\). The other cases for when we have \(\varphi_2\) is similar. The north and south pole of the circle are mapped via \(\psi_2 \circ F \circ \phi_2\) and \(\psi_1 \circ F \circ \phi_1\) respectively. So \(F\) is a bijection between between the two charts of the manifolds and the inverse mapping between the charts is a smooth function in all its charts as well \((\psi_i \circ F \circ \varphi_j)^{-1}\) . So it is a diffeomorphism.

Problem 3: Let \(G\) be a Lie Group. Let \(G_0\) be the connected component of \(G\) containing the identity. Prove that every connected component of \(G\) is diffeomorphic to \(G_0\). (I'm using the definition that a connected component of a topological space is a maximal connected subset.)

Proof: If \(H \subseteq G\) is a connected component and \(h \in H\). Consider the left action \(L_h : G \rightarrow G\) were \(L_h(x)=hx\). Then \(L_h(1)=h\). Since \(L_h\) is a diffeomorphism, connected components are mapped to connected components. Then \(hG_0 = H\).

Problem 4: If \(f: M \rightarrow N\) is a smooth map and \(\vec{v} \in T_pM\) is represented by a curve \(\gamma:(- \epsilon, \epsilon) \rightarrow M\), we defined \(d_pf(\vec{v}) \in T_{f(p)}N\) as the curve \(f \circ \gamma : (- \epsilon , \epsilon) \rightarrow N\)

part a: If \(\gamma_1 \sim \gamma_2\), then \(d_pf(\gamma_1) \sim d_pf(\gamma_2)\). (Where \(\gamma_1 \sim \gamma_2\) if for any chart \((U, \varphi)\) around \(p \in M\), we have that \((\varphi \circ \gamma_1)'(0) = (\varphi \circ \gamma_2)'(0)\) )

Solution: Let \((U, \varphi)\) be a chart on \(N\) and let \(p \in U\) such that \(\gamma_i(0) = p , i=1,2\). Observe that \((\varphi \circ (f \circ \gamma_1))'(0) = \varphi(f \circ \gamma_1(0))(f \circ \gamma_1)'(0) = \varphi(f(p))(f \circ \gamma_1)'(0)\). Using the fact that \(\gamma_1 \sim \gamma_2 \Rightarrow \gamma_1'(0) = \gamma_2'(0)\). It follows that \(\varphi(f(p))(f \circ \gamma_1)'(0) = \varphi(f(p))f(\gamma_1(0))\gamma_1'(0) = \varphi(f(p))f(\gamma_2(0))\gamma_2'(0) = \varphi(f(p))(f \circ \gamma_2)'(0) = \varphi(f \circ \gamma_2(0))(f \circ \gamma_2)'(0) = (\varphi \circ (f \circ \gamma_2))'(0)\). Thus \(d_pf(\gamma_1) \sim d_pf(\gamma_2)\)

part b: Suppose \((U , \varphi)\) is a chart around \(p\) with coordinate functions, \(x^i\), and \(\vec{v}\) is represented in those coordinates as \(\vec{v} = v^i \frac{\partial}{\partial x^i}\). If \((V, \psi)\) is a chart aroudn \(f(p)\) with coordinate functions \(y^j\), what is the representation of \(d_pf(\vec{v})\) in the \(y^j\) coordinates?

Solution: Given that \(\vec{v} = v^i \frac{\partial}{\partial x^i}\). We see that \(d_pf(\vec{v}) = f(v)^j \frac{\partial}{\partial y^j}\). The derivative of \(f(v)\) in the \(y^j \) direction.

part c: If \(D_{\vec{v}}\) is the derivation at \(p\) associated to \(\vec{v} \in T_pM\). What is the derivation at \(f(p)\) associated to \(d_pf(\vec{v})\)? Describe what the derivation does to functions \(g\) defined in a neighborhood of \(f(p) \in N\). (Don't need to appeal to coordinate descriptions)

Solution: In words, the derivation associated to \(D_{d_pf(v)}|_f(p)\) of a function \(g\) is the derivative of \(g\) at \(f(p)\) in the \(y^j\) direction scaled by \(f(v)^j\). i.e.\(D_{d_pf(v)}|_{f(p)} (g) = f(v)^j \frac{\partial g}{\partial y^j} f(p) \) Sorry, this appeals to local coordinates. But this is what made sense in my head.

Problem 5: Let \(F: M \rightarrow N\) and \(G: N \rightarrow P\) be smooth maps, and let \(p \in M\).

part a: \(F_*: T_pM \rightarrow T_{F(p)}N\) is linear.

Solution: Let \(f,g \in T_pM, h \in C^{\infty}N\). Then \(F_*(f+g)(h) = (f+g)(h \circ F)= f(h \circ F) + g(h \circ F) = F_*(f)(h) + F_*(g)(h)\). Now let \(c\ in \mathbb{R}, f \in T_pM, h \in C^{\infty}N\). Then \(F_*(cf)(h) = (cf)(h \circ F) = c \cdot f(h \circ F) = c F_*(f)(h)\). Thus, \(F_*\) is linear.

part b: \((G \circ F)_* = G_* \circ F_*: T_pM \rightarrow T_{G \circ F(p)}P\)

Solution: Let \(f\in T_pM, h \in C^{\infty}P\). Then \((G \circ F)_*(f)(h) = f(h \circ (G \circ F)) = f((h \circ G) \circ F) = F_*(f)(h \circ G) = G_*(F_*(f)(h)) = G_* \circ F_*(f)(h)\).

part c: \((Id_M)_* = Id_{T_pM}: T_pM \rightarrow T_pM\)

Solution: Let \(v \in T_pM, f \in C^{\infty}M\). Then \(Id_*(v)(f) = v(f \circ Id) = v(f)\). Whereas \(Id_{T_pM}:T_pM \rightarrow T_pM\) sends \(v\) to itself. So \(v\) as a derivation at \(p\) acts identically on \(f\). Hence, equality between the maps.

part d: If \(F\) is a diffeomorphism, then \(F_*: T_pM \rightarrow T_{F(p)}N\) is an isomoprhism.

Solution: Let \(v\in T_pM\) and \(f \in C^{\infty}M\) If \(F\) is a diffeomorphism. Then we get that \(F^{-1}: T_{f(p)}N \rightarrow T_pM\) is a smooth bijection. So \(F_*^{-1} \circ F_*(v) = F_*^{-1}(v(f \circ F)) = v ((f \circ F) \circ F^{-1}) = v(f)\). Which is the identity! So \(F_*\) is a bijection and linear by our previous result. So we conclude that this map is an isomorphism.

Problem 6: Let \(G\) be a Lie group.

part a: Show that \(m_*: T_eG \oplus T_eG\) is given by \(m_*(X,Y) = X+Y\) where \(m\) is the multiplcation map.

Solution: Observe that \(m_{(e,e)}(X,0) = m_e(X) = m(X,e) = X = Id(X)\) and \(m_{(e,e)}(0,Y) = Y = Id(Y)\). So really, \(m_*(X,0)\) is the differential of the identity map, which would return us our derivation. We then obtain that \(m_*(X,Y) = m_*(X,0) + m_*(0,Y) = X +Y\) by our previous equalities.

part b: Show that \(i_*:T_eG \rightarrow T_eG\) is given by \(i_*X = -X\) where \(i\) is the inversion map.

Solution: Consider the maps \(f: G \rightarrow G \times G, f(x) = (x,x) , g: G\times G \rightarrow G \times G, g(x,y) = (x, y^{-1}), m:G\times G \rightarrow G, m(x,y) = xy\). Note that the composition \(c = m \circ g \circ f\) is the constant map to the identity. The derivation of a constant map is zero. No change! Thus \(0 = c_*(X) = m_*(g_*(f_*(X))) = m_*(g_*(X,X)) = m_*(X, i_*(X)) = X +i_*(X)\). Thus, \(i_*(X) = - X\). The last equality follows from our previous result and the one before that follows from the fact that \(g\) is the identity and the inverse componentwise.