Talk:Normal maps - (non)-examples (Ex)

1) First note that, for any closed manifold $M$$<wikitex>; 1) First note that, for any closed manifold $M$, there is a map $f:M \to S^n$ giving by contracting the complement of an embedded disc $D^n \subseteq M$. At a point $x \in D^n$, this map is bijective and has local degree $\deg(f)_x=1$ and so $\deg(f)=1$ by the fact that the degree is the sum of local degrees, i.e. $\deg(f) = \sum_{i=1}^r \deg(f)_{x_i}$ where $y=f(x)$ and $f^{-1}(y)=\{x_1, \cdots, x_r\}$ which in this case is just $\{x\}$. Now specialise to the case $n=4 $ and $M= \CP^2$, $X=S^4$ and consider $\xi=(\R^4 \to E \to S^4)$, an arbitrary 4-dimensional real vector bundle over $S^4$. If there is a normal map $(f,\bar{f}): \nu_M \to \xi$ then, since this is a fibrewise isomorphism we must have that $\nu_M \cong (\bar{f})^*(\xi)$. Consider the first Pontryagin classes $p_1(\nu_M) \in H^4(M)$ and $p_1(\xi) \in H^4(S^4)$. With respect to some identifications $H^4(M)=\mathbb{Z}$ and $H^4(S^4)=\mathbb{Z}$, we will show that $p_1(T M) = 3$ and $p_1(\xi) = 2k$ for some $k \in \Z$ which will depend on the choice of bundle $\xi$. This would complete the proof since this would imply that $p_1(\nu_M)=-3$ but also that $p_1(\nu_M) = (\bar{f})^*(2k) = 2 \cdot (\bar{f})^*(k)$ which is divisible by two. Note that that fact that $(f,\bar{f}): \nu_M \to \xi$ is only a stable bundle map is dealt with here by the fact that $p_1$ is a stable bundle invariant and so gives that same construction for any two stably isomorphic bundles of the same dimension. To show that $p_1(T M)=3$ for $M = \CP^2$ and some choice of identification $H^4(M)=\Z$, we can use a number of techniques. One quick trick is to note that the Hirzebruch Signature Theorem applies to $M$ and gives that $\sigma(M) =\frac{1}{3} p_1(TM)[M]$ by the fact that the first polynomial is the multiplicative sequence corresponding to the $L$-genus is $L_1 = \frac{1}{3} p_1$. Now $H^2(\CP^2)=\Z$ and the intersection form is $(1)$ gives that $\sigma(\CP^2)=1$ and so $p_1(TM)[M] =3$. Hence $p_1(TM)=3$ with appropriate identifications. Another, more elementary, method is to show that $T \CP^2 \oplus \tau \cong \gamma_{\mathbb{C}}^2 \oplus \gamma_{\mathbb{C}}^2$ where $\gamma_{\mathbb{C}}^2$ is the tautological complex line bundle over $\CP^2$. This allows us to compute the total Chern class immediately which can then be used to recover the Pontryagin classes using the the highly useful formula M$, there is a map $f:M \to S^n$$f:M \to S^n$ giving by contracting the complement of an embedded disc $D^n \subseteq M$$D^n \subseteq M$. At a point $x \in D^n$$x \in D^n$, this map is bijective and has local degree $\deg(f)_x=1$$\deg(f)_x=1$ and so $\deg(f)=1$$\deg(f)=1$ by the fact that the degree is the sum of local degrees, i.e. $\deg(f) = \sum_{i=1}^r \deg(f)_{x_i}$$\deg(f) = \sum_{i=1}^r \deg(f)_{x_i}$ where $y=f(x)$$y=f(x)$ and $f^{-1}(y)=\{x_1, \cdots, x_r\}$$f^{-1}(y)=\{x_1, \cdots, x_r\}$ which in this case is just $\{x\}$$\{x\}$.

Now specialise to the case $n=4$$n=4$ and $M= \CP^2$$M= \CP^2$, $X=S^4$$X=S^4$ and consider $\xi=(\R^4 \to E \to S^4)$$\xi=(\R^4 \to E \to S^4)$, an arbitrary 4-dimensional real vector bundle over $S^4$$S^4$. If there is a normal map $(f,\bar{f}): \nu_M \to \xi$$(f,\bar{f}): \nu_M \to \xi$ then, since this is a fibrewise isomorphism we must have that $\nu_M \cong (\bar{f})^*(\xi)$$\nu_M \cong (\bar{f})^*(\xi)$. Consider the first Pontryagin classes $p_1(\nu_M) \in H^4(M)$$p_1(\nu_M) \in H^4(M)$ and $p_1(\xi) \in H^4(S^4)$$p_1(\xi) \in H^4(S^4)$. With respect to some identifications $H^4(M)=\mathbb{Z}$$H^4(M)=\mathbb{Z}$ and $H^4(S^4)=\mathbb{Z}$$H^4(S^4)=\mathbb{Z}$, we will show that $p_1(T M) = 3$$p_1(T M) = 3$ and $p_1(\xi) = 2k$$p_1(\xi) = 2k$ for some $k \in \Z$$k \in \Z$ which will depend on the choice of bundle $\xi$$\xi$.

This would complete the proof since this would imply that $p_1(\nu_M)=-3$$p_1(\nu_M)=-3$ but also that $p_1(\nu_M) = (\bar{f})^*(2k) = 2 \cdot (\bar{f})^*(k)$$p_1(\nu_M) = (\bar{f})^*(2k) = 2 \cdot (\bar{f})^*(k)$ which is divisible by two. Note that that fact that $(f,\bar{f}): \nu_M \to \xi$$(f,\bar{f}): \nu_M \to \xi$ is only a stable bundle map is dealt with here by the fact that $p_1$$p_1$ is a stable bundle invariant and so gives that same construction for any two stably isomorphic bundles of the same dimension.

To show that $p_1(T M)=3$$p_1(T M)=3$ for $M = \CP^2$$M = \CP^2$ and some choice of identification $H^4(M)=\Z$$H^4(M)=\Z$, we can use a number of techniques. One quick trick is to note that the Hirzebruch Signature Theorem applies to $M$$M$ and gives that $\sigma(M) =\frac{1}{3} p_1(TM)[M]$$\sigma(M) =\frac{1}{3} p_1(TM)[M]$ by the fact that the first polynomial is the multiplicative sequence corresponding to the $L$$L$-genus is $L_1 = \frac{1}{3} p_1$$L_1 = \frac{1}{3} p_1$. Now $H^2(\CP^2)=\Z$$H^2(\CP^2)=\Z$ and the intersection form is $(1)$$(1)$ gives that $\sigma(\CP^2)=1$$\sigma(\CP^2)=1$ and so $p_1(TM)[M] =3$$p_1(TM)[M] =3$. Hence $p_1(TM)=3$$p_1(TM)=3$ with appropriate identifications. Another, more elementary, method is to show that $T \CP^2 \oplus \tau \cong \gamma_{\mathbb{C}}^2 \oplus \gamma_{\mathbb{C}}^2$$T \CP^2 \oplus \tau \cong \gamma_{\mathbb{C}}^2 \oplus \gamma_{\mathbb{C}}^2$ where $\gamma_{\mathbb{C}}^2$$\gamma_{\mathbb{C}}^2$ is the tautological complex line bundle over $\CP^2$$\CP^2$. This allows us to compute the total Chern class immediately which can then be used to recover the Pontryagin classes using the the highly useful formula $1-p_1(\xi_\mathbb{R})+p_2(\xi_\mathbb{R})-\cdots = (1-c_1(\xi)+c_2(\xi)-\cdots)(1+c_1(\xi)+c_2(\xi)+\cdots)$$1-p_1(\xi_\mathbb{R})+p_2(\xi_\mathbb{R})-\cdots = (1-c_1(\xi)+c_2(\xi)-\cdots)(1+c_1(\xi)+c_2(\xi)+\cdots)$ where $\xi_\mathbb{R}$$\xi_\mathbb{R}$ denotes that fibrewise replacement of $\C$$\C$ with $\R^2$$\R^2$.

To show that $p_1(\xi)$$p_1(\xi)$ is always even is more tricky, and is asserted (but not proven) in Milnor's paper which proves the existence of exotic $7$$7$-spheres. Firstly let $\mathbb{S}(\xi) = (S^3 \to \mathbb{S}(E) \to S^4)$$\mathbb{S}(\xi) = (S^3 \to \mathbb{S}(E) \to S^4)$ be the sphere bundle corresponding to $\xi$$\xi$. Recall that, by clutching construction for bundles over spheres, there is a 1-1 correspondence between fibre bundles over $S^n$$S^n$ with fixed fibre $F$$F$ and structure group $G$$G$ and $\pi_{n-1}(G)$$\pi_{n-1}(G)$. Hence oriented $S^3$$S^3$-bundles over $S^4$$S^4$ are classified by $\pi_3(SO(4))$$\pi_3(SO(4))$. One can show explicitly that $SO(4) \cong S^3 \times SO(3)$$SO(4) \cong S^3 \times SO(3)$ and so has universal cover $S^3 \times S^3$$S^3 \times S^3$ and therefore $\pi_3(SO(4)) \cong \Z^2$$\pi_3(SO(4)) \cong \Z^2$ since covering maps induce isomorphisms on higher homotopy groups. Fix an explicit identification of $\Z^2$$\Z^2$ with $\pi_3(SO(4))$$\pi_3(SO(4))$. Therefore $\mathbb{S}(E) = E_{i,j}$$\mathbb{S}(E) = E_{i,j}$ where $(i,j) \in \Z^2$$(i,j) \in \Z^2$ labels the element of $\pi_3(SO(4))$$\pi_3(SO(4))$ corresponding to the sphere bundle.

There is a map $\Z^2 \to H^4(S^4)=\Z$$\Z^2 \to H^4(S^4)=\Z$ given by mapping $(i,j)$$(i,j)$ to $p_1(E_{i,j})$$p_1(E_{i,j})$. By considering the inclusion of the fibre of the corresponding real vector bundle into the total space, one can show that this map factors as group homomorphisms $\Z^2 = \pi_3(SO(4)) \to \pi_4(G_4(\mathbb{R}^8) \to H^4(S^4) = \Z$$\Z^2 = \pi_3(SO(4)) \to \pi_4(G_4(\mathbb{R}^8) \to H^4(S^4) = \Z$ where $G_4(\mathbb{R}^8)$$G_4(\mathbb{R}^8)$ is the Grassmannian , i.e. the $4$$4$-dimensional linear subspaces of $\mathbb{R}^8$$\mathbb{R}^8$. Hence $p_1(E_{i,j})$$p_1(E_{i,j})$ is linear in $i$$i$ and $j$$j$ and so of the form $ai+bj$$ai+bj$. By considering the quaternionic conjugate on the fibre, we can also show that $p_1(E_{i,j})=p_1(E_{-j,-i})$$p_1(E_{i,j})=p_1(E_{-j,-i})$ which guarantees that $p_1(E_{i,j})=a(i-j)$$p_1(E_{i,j})=a(i-j)$ for some $a \in \Z$$a \in \Z$ and some identification $H^4(S^4)=\mathbb{Z}$$H^4(S^4)=\mathbb{Z}$. To finish, we can check that $E_{0,1}$$E_{0,1}$ is the tautological bundle over $\HP^1$$\HP^1$ by using an explicit identification of $\Z^2$$\Z^2$ with $\pi_3(SO(4))$$\pi_3(SO(4))$. This shows that $p_1(E_01)=-2 \eta$$p_1(E_01)=-2 \eta$ where $\eta$$\eta$ is the pullback of the chosen generator of $H^4(S^4)$$H^4(S^4)$ along an identification $\mathbb{H}P^1 \cong S^4$$\mathbb{H}P^1 \cong S^4$. This shows that $a=2$$a=2$ and hence $p_(E_{i,j})=2(I-j)$$p_(E_{i,j})=2(I-j)$ which completes the proof.

See also Obstruction classes and Pontrjagin classes (Ex).

2) For any $n$$n$, let $M=X=S^n$$M=X=S^n$. Note that the stable normal bundle $\nu_S^n$$\nu_S^n$ is trivial since the normal bundle of the codimension one embedding $S^n \hookrightarrow \mathbb{R}^{n+1}$$S^n \hookrightarrow \mathbb{R}^{n+1}$ is trivial. Hence any map $f: S^n \to S^n$$f: S^n \to S^n$ extends to a bundle map $(f,\bar{f}): \nu_{S^n} \to \nu_{S^n}$$(f,\bar{f}): \nu_{S^n} \to \nu_{S^n}$ and so it suffices to find degree $d$$d$ maps $f_d: S^n \to S^n$$f_d: S^n \to S^n$ for any $d$$d$.

For $n=1$$n=1$, a degree $d$$d$ map $f_d: S^1 \to S^1$$f_d: S^1 \to S^1$ is given by wrapping the circle $d$$d$ times around itself. By suspending $n-1$$n-1$ times we get a map $\Sigma^{n-1}(f_d): S^n= \Sigma^{n-1}(S^1) \to \Sigma^{n-1}(S^1) = S^n$$\Sigma^{n-1}(f_d): S^n= \Sigma^{n-1}(S^1) \to \Sigma^{n-1}(S^1) = S^n$ which we can show still has degree $d$$d$ by using local degrees as in the previous exercise.

3) We will show that there is a degree one normal map $(f, \bar{f}): F_g \to F_{g'}$$(f, \bar{f}): F_g \to F_{g'}$ if and only if $g \ge g'$$g \ge g'$. Firstly note that, as above, the codimension one embedding $F_g \hookrightarrow \mathbb{R}^3$$F_g \hookrightarrow \mathbb{R}^3$ is trivial and so $\nu_{F_g}$$\nu_{F_g}$ is trivial for all $g$$g$. Hence any map extends to a normal map and it remains to show that there is a degree one map $f: F_g \to F_{g'}$$f: F_g \to F_{g'}$ if and only if $g \ge g'$$g \ge g'$.

If $g \ge g'$$g \ge g'$, then we can get a map $f: F_g \to F_{g'}$$f: F_g \to F_{g'}$ by contracting the complement of the standard $F_{g'} \setminus D^n \subseteq F_g$$F_{g'} \setminus D^n \subseteq F_g$. This is degree one using local degree as in the first exercise.

Let $g < g'$$g < g'$ and suppose there is a map $f^*: H^*(F_{g'}) \to H^*(F_g)$$f^*: H^*(F_{g'}) \to H^*(F_g)$ of cohomology rings sending the generator of $H^2(F_g)$$H^2(F_g)$ to the generator of $H^2(F_{g'})$$H^2(F_{g'})$. Note that $H^*(F_g)$$H^*(F_g)$ is generated by $\alpha_1, \cdots, \alpha_g, \beta_1, \cdots, \beta_g$$\alpha_1, \cdots, \alpha_g, \beta_1, \cdots, \beta_g$ in degree one and $\gamma$$\gamma$ is degree two such that $\gamma= \alpha_i \cup \beta_i$$\gamma= \alpha_i \cup \beta_i$ for $i=1, \cdots, g$$i=1, \cdots, g$ and all other products of the degree one generators zero. This can be lifted from the quotient map $F_g \to \bigvee_{i=1}^g F_1$$F_g \to \bigvee_{i=1}^g F_1$ and the structure of the cohomology ring of $\bigvee_{i=1}^g F_1$$\bigvee_{i=1}^g F_1$ coming from that of $F_1$$F_1$. Since $g < g'$$g < g'$, the map $f^*: \mathbb{Z}^{2g'} = H^1(F_{g'}) \to H^1(F_g)=\mathbb{Z}^{2g}$$f^*: \mathbb{Z}^{2g'} = H^1(F_{g'}) \to H^1(F_g)=\mathbb{Z}^{2g}$ has a non-zero element $\mu= \sum_{i=1}^g (a_i \alpha_i + b_i \beta_i)$$\mu= \sum_{i=1}^g (a_i \alpha_i + b_i \beta_i)$ in its kernel. Suppose, without loss of generality, that $a_j \ne 0$$a_j \ne 0$. Then $\mu \cup \alpha_j = a_j \gamma$$\mu \cup \alpha_j = a_j \gamma$ and so $f^*(\mu \cup \alpha_j) \ne 0 \in H^2(F_g)$$f^*(\mu \cup \alpha_j) \ne 0 \in H^2(F_g)$. However $f^*(\mu \cup \alpha_j) = f^*(\mu) \cup f*(\alpha_j)$$f^*(\mu \cup \alpha_j) = f^*(\mu) \cup f*(\alpha_j)$ and $f^*(\mu)=0$$f^*(\mu)=0$ by hypothesis. This is a contradiction.

This can also be proven by using that a degree one map $f: F_g \to F_{g'}$$f: F_g \to F_{g'}$ induces a surjection $f_*: \pi_1(F_g) \to \pi_1(F_{g'})$$f_*: \pi_1(F_g) \to \pi_1(F_{g'})$ and so a surjection $\Z^{2g} = \pi_1(F_g)^{ab} \to \pi_1(F_{g'})^{ab} = \Z^{2g'}$$\Z^{2g} = \pi_1(F_g)^{ab} \to \pi_1(F_{g'})^{ab} = \Z^{2g'}$ which gives a contradiction as before.

-p_1(\xi_\mathbb{R})+p_2(\xi_\mathbb{R})-\cdots = (1-c_1(\xi)+c_2(\xi)-\cdots)(1+c_1(\xi)+c_2(\xi)+\cdots)$ where $\xi_\mathbb{R}$ denotes that fibrewise replacement of $\C$ with $\R^2$. To show that $p_1(\xi)$ is always even is more tricky, and is asserted (but not proven) in Milnor's paper which proves the existence of exotic $-spheres. Firstly let $\mathbb{S}(\xi) = (S^3 \to \mathbb{S}(E) \to S^4)$ be the sphere bundle corresponding to $\xi$. Recall that, by clutching construction for bundles over spheres, there is a 1-1 correspondence between fibre bundles over $S^n$ with fixed fibre $F$ and structure group $G$ and $\pi_{n-1}(G)$. Hence oriented $S^3$-bundles over $S^4$ are classified by $\pi_3(SO(4))$. One can show explicitly that $SO(4) \cong S^3 \times SO(3)$ and so has universal cover $S^3 \times S^3$ and therefore $\pi_3(SO(4)) \cong \Z^2$ since covering maps induce isomorphisms on higher homotopy groups. Fix an explicit identification of $\Z^2$ with $\pi_3(SO(4))$. Therefore $\mathbb{S}(E) = E_{i,j}$ where $(i,j) \in \Z^2$ labels the element of $\pi_3(SO(4))$ corresponding to the sphere bundle. There is a map $\Z^2 \to H^4(S^4)=\Z$ given by mapping $(i,j)$ to $p_1(E_{i,j})$. By considering the inclusion of the fibre of the corresponding real vector bundle into the total space, one can show that this map factors as group homomorphisms $\Z^2 = \pi_3(SO(4)) \to \pi_4(G_4(\mathbb{R}^8) \to H^4(S^4) = \Z$ where $G_4(\mathbb{R}^8)$ is the Grassmannian , i.e. the $-dimensional linear subspaces of $\mathbb{R}^8$. Hence $p_1(E_{i,j})$ is linear in $i$ and $j$ and so of the form $ai+bj$. By considering the quaternionic conjugate on the fibre, we can also show that $p_1(E_{i,j})=p_1(E_{-j,-i})$ which guarantees that $p_1(E_{i,j})=a(i-j)$ for some $a \in \Z$ and some identification $H^4(S^4)=\mathbb{Z}$. To finish, we can check that $E_{0,1}$ is the tautological bundle over $\HP^1$ by using an explicit identification of $\Z^2$ with $\pi_3(SO(4))$. This shows that $p_1(E_01)=-2 \eta$ where $\eta$ is the pullback of the chosen generator of $H^4(S^4)$ along an identification $\mathbb{H}P^1 \cong S^4$. This shows that $a=2$ and hence $p_(E_{i,j})=2(I-j)$ which completes the proof. See also [[Obstruction classes and Pontrjagin classes (Ex)]]. 2) For any $n$, let $M=X=S^n$. Note that the stable normal bundle $\nu_S^n$ is trivial since the normal bundle of the codimension one embedding $S^n \hookrightarrow \mathbb{R}^{n+1}$ is trivial. Hence any map $f: S^n \to S^n$ extends to a bundle map $(f,\bar{f}): \nu_{S^n} \to \nu_{S^n}$ and so it suffices to find degree $d$ maps $f_d: S^n \to S^n$ for any $d$. For $n=1$, a degree $d$ map $f_d: S^1 \to S^1$ is given by wrapping the circle $d$ times around itself. By suspending $n-1$ times we get a map $\Sigma^{n-1}(f_d): S^n= \Sigma^{n-1}(S^1) \to \Sigma^{n-1}(S^1) = S^n$ which we can show still has degree $d$ by using local degrees as in the previous exercise. 3) We will show that there is a degree one normal map $(f, \bar{f}): F_g \to F_{g'}$ if and only if $g \ge g'$. Firstly note that, as above, the codimension one embedding $F_g \hookrightarrow \mathbb{R}^3$ is trivial and so $\nu_{F_g}$ is trivial for all $g$. Hence any map extends to a normal map and it remains to show that there is a degree one map $f: F_g \to F_{g'}$ if and only if $g \ge g'$. If $g \ge g'$, then we can get a map $f: F_g \to F_{g'}$ by contracting the complement of the standard $F_{g'} \setminus D^n \subseteq F_g$. This is degree one using local degree as in the first exercise. Let $g < g'$ and suppose there is a map $f^*: H^*(F_{g'}) \to H^*(F_g)$ of cohomology rings sending the generator of $H^2(F_g)$ to the generator of $H^2(F_{g'})$. Note that $H^*(F_g)$ is generated by $\alpha_1, \cdots, \alpha_g, \beta_1, \cdots, \beta_g$ in degree one and $\gamma$ is degree two such that $\gamma= \alpha_i \cup \beta_i$ for $i=1, \cdots, g$ and all other products of the degree one generators zero. This can be lifted from the quotient map $F_g \to \bigvee_{i=1}^g F_1$ and the structure of the cohomology ring of $\bigvee_{i=1}^g F_1$ coming from that of $F_1$. Since $g < g'$, the map $f^*: \mathbb{Z}^{2g'} = H^1(F_{g'}) \to H^1(F_g)=\mathbb{Z}^{2g}$ has a non-zero element $\mu= \sum_{i=1}^g (a_i \alpha_i + b_i \beta_i)$ in its kernel. Suppose, without loss of generality, that $a_j \ne 0$. Then $\mu \cup \alpha_j = a_j \gamma$ and so $f^*(\mu \cup \alpha_j) \ne 0 \in H^2(F_g)$. However $f^*(\mu \cup \alpha_j) = f^*(\mu) \cup f*(\alpha_j)$ and $f^*(\mu)=0$ by hypothesis. This is a contradiction. This can also be proven by using that a degree one map $f: F_g \to F_{g'}$ induces a surjection $f_*: \pi_1(F_g) \to \pi_1(F_{g'})$ and so a surjection $\Z^{2g} = \pi_1(F_g)^{ab} \to \pi_1(F_{g'})^{ab} = \Z^{2g'}$ which gives a contradiction as before. [[Category:Exercises]] [[Category:Exercises with solution]]M, there is a map $f:M \to S^n$$f:M \to S^n$ giving by contracting the complement of an embedded disc $D^n \subseteq M$$D^n \subseteq M$. At a point $x \in D^n$$x \in D^n$, this map is bijective and has local degree $\deg(f)_x=1$$\deg(f)_x=1$ and so $\deg(f)=1$$\deg(f)=1$ by the fact that the degree is the sum of local degrees, i.e. $\deg(f) = \sum_{i=1}^r \deg(f)_{x_i}$$\deg(f) = \sum_{i=1}^r \deg(f)_{x_i}$ where $y=f(x)$$y=f(x)$ and $f^{-1}(y)=\{x_1, \cdots, x_r\}$$f^{-1}(y)=\{x_1, \cdots, x_r\}$ which in this case is just $\{x\}$$\{x\}$.

Now specialise to the case $n=4$$n=4$ and $M= \CP^2$$M= \CP^2$, $X=S^4$$X=S^4$ and consider $\xi=(\R^4 \to E \to S^4)$$\xi=(\R^4 \to E \to S^4)$, an arbitrary 4-dimensional real vector bundle over $S^4$$S^4$. If there is a normal map $(f,\bar{f}): \nu_M \to \xi$$(f,\bar{f}): \nu_M \to \xi$ then, since this is a fibrewise isomorphism we must have that $\nu_M \cong (\bar{f})^*(\xi)$$\nu_M \cong (\bar{f})^*(\xi)$. Consider the first Pontryagin classes $p_1(\nu_M) \in H^4(M)$$p_1(\nu_M) \in H^4(M)$ and $p_1(\xi) \in H^4(S^4)$$p_1(\xi) \in H^4(S^4)$. With respect to some identifications $H^4(M)=\mathbb{Z}$$H^4(M)=\mathbb{Z}$ and $H^4(S^4)=\mathbb{Z}$$H^4(S^4)=\mathbb{Z}$, we will show that $p_1(T M) = 3$$p_1(T M) = 3$ and $p_1(\xi) = 2k$$p_1(\xi) = 2k$ for some $k \in \Z$$k \in \Z$ which will depend on the choice of bundle $\xi$$\xi$.

This would complete the proof since this would imply that $p_1(\nu_M)=-3$$p_1(\nu_M)=-3$ but also that $p_1(\nu_M) = (\bar{f})^*(2k) = 2 \cdot (\bar{f})^*(k)$$p_1(\nu_M) = (\bar{f})^*(2k) = 2 \cdot (\bar{f})^*(k)$ which is divisible by two. Note that that fact that $(f,\bar{f}): \nu_M \to \xi$$(f,\bar{f}): \nu_M \to \xi$ is only a stable bundle map is dealt with here by the fact that $p_1$$p_1$ is a stable bundle invariant and so gives that same construction for any two stably isomorphic bundles of the same dimension.

To show that $p_1(T M)=3$$p_1(T M)=3$ for $M = \CP^2$$M = \CP^2$ and some choice of identification $H^4(M)=\Z$$H^4(M)=\Z$, we can use a number of techniques. One quick trick is to note that the Hirzebruch Signature Theorem applies to $M$$M$ and gives that $\sigma(M) =\frac{1}{3} p_1(TM)[M]$$\sigma(M) =\frac{1}{3} p_1(TM)[M]$ by the fact that the first polynomial is the multiplicative sequence corresponding to the $L$$L$-genus is $L_1 = \frac{1}{3} p_1$$L_1 = \frac{1}{3} p_1$. Now $H^2(\CP^2)=\Z$$H^2(\CP^2)=\Z$ and the intersection form is $(1)$$(1)$ gives that $\sigma(\CP^2)=1$$\sigma(\CP^2)=1$ and so $p_1(TM)[M] =3$$p_1(TM)[M] =3$. Hence $p_1(TM)=3$$p_1(TM)=3$ with appropriate identifications. Another, more elementary, method is to show that $T \CP^2 \oplus \tau \cong \gamma_{\mathbb{C}}^2 \oplus \gamma_{\mathbb{C}}^2$$T \CP^2 \oplus \tau \cong \gamma_{\mathbb{C}}^2 \oplus \gamma_{\mathbb{C}}^2$ where $\gamma_{\mathbb{C}}^2$$\gamma_{\mathbb{C}}^2$ is the tautological complex line bundle over $\CP^2$$\CP^2$. This allows us to compute the total Chern class immediately which can then be used to recover the Pontryagin classes using the the highly useful formula $1-p_1(\xi_\mathbb{R})+p_2(\xi_\mathbb{R})-\cdots = (1-c_1(\xi)+c_2(\xi)-\cdots)(1+c_1(\xi)+c_2(\xi)+\cdots)$$1-p_1(\xi_\mathbb{R})+p_2(\xi_\mathbb{R})-\cdots = (1-c_1(\xi)+c_2(\xi)-\cdots)(1+c_1(\xi)+c_2(\xi)+\cdots)$ where $\xi_\mathbb{R}$$\xi_\mathbb{R}$ denotes that fibrewise replacement of $\C$$\C$ with $\R^2$$\R^2$.

To show that $p_1(\xi)$$p_1(\xi)$ is always even is more tricky, and is asserted (but not proven) in Milnor's paper which proves the existence of exotic $7$$7$-spheres. Firstly let $\mathbb{S}(\xi) = (S^3 \to \mathbb{S}(E) \to S^4)$$\mathbb{S}(\xi) = (S^3 \to \mathbb{S}(E) \to S^4)$ be the sphere bundle corresponding to $\xi$$\xi$. Recall that, by clutching construction for bundles over spheres, there is a 1-1 correspondence between fibre bundles over $S^n$$S^n$ with fixed fibre $F$$F$ and structure group $G$$G$ and $\pi_{n-1}(G)$$\pi_{n-1}(G)$. Hence oriented $S^3$$S^3$-bundles over $S^4$$S^4$ are classified by $\pi_3(SO(4))$$\pi_3(SO(4))$. One can show explicitly that $SO(4) \cong S^3 \times SO(3)$$SO(4) \cong S^3 \times SO(3)$ and so has universal cover $S^3 \times S^3$$S^3 \times S^3$ and therefore $\pi_3(SO(4)) \cong \Z^2$$\pi_3(SO(4)) \cong \Z^2$ since covering maps induce isomorphisms on higher homotopy groups. Fix an explicit identification of $\Z^2$$\Z^2$ with $\pi_3(SO(4))$$\pi_3(SO(4))$. Therefore $\mathbb{S}(E) = E_{i,j}$$\mathbb{S}(E) = E_{i,j}$ where $(i,j) \in \Z^2$$(i,j) \in \Z^2$ labels the element of $\pi_3(SO(4))$$\pi_3(SO(4))$ corresponding to the sphere bundle.

There is a map $\Z^2 \to H^4(S^4)=\Z$$\Z^2 \to H^4(S^4)=\Z$ given by mapping $(i,j)$$(i,j)$ to $p_1(E_{i,j})$$p_1(E_{i,j})$. By considering the inclusion of the fibre of the corresponding real vector bundle into the total space, one can show that this map factors as group homomorphisms $\Z^2 = \pi_3(SO(4)) \to \pi_4(G_4(\mathbb{R}^8) \to H^4(S^4) = \Z$$\Z^2 = \pi_3(SO(4)) \to \pi_4(G_4(\mathbb{R}^8) \to H^4(S^4) = \Z$ where $G_4(\mathbb{R}^8)$$G_4(\mathbb{R}^8)$ is the Grassmannian , i.e. the $4$$4$-dimensional linear subspaces of $\mathbb{R}^8$$\mathbb{R}^8$. Hence $p_1(E_{i,j})$$p_1(E_{i,j})$ is linear in $i$$i$ and $j$$j$ and so of the form $ai+bj$$ai+bj$. By considering the quaternionic conjugate on the fibre, we can also show that $p_1(E_{i,j})=p_1(E_{-j,-i})$$p_1(E_{i,j})=p_1(E_{-j,-i})$ which guarantees that $p_1(E_{i,j})=a(i-j)$$p_1(E_{i,j})=a(i-j)$ for some $a \in \Z$$a \in \Z$ and some identification $H^4(S^4)=\mathbb{Z}$$H^4(S^4)=\mathbb{Z}$. To finish, we can check that $E_{0,1}$$E_{0,1}$ is the tautological bundle over $\HP^1$$\HP^1$ by using an explicit identification of $\Z^2$$\Z^2$ with $\pi_3(SO(4))$$\pi_3(SO(4))$. This shows that $p_1(E_01)=-2 \eta$$p_1(E_01)=-2 \eta$ where $\eta$$\eta$ is the pullback of the chosen generator of $H^4(S^4)$$H^4(S^4)$ along an identification $\mathbb{H}P^1 \cong S^4$$\mathbb{H}P^1 \cong S^4$. This shows that $a=2$$a=2$ and hence $p_(E_{i,j})=2(I-j)$$p_(E_{i,j})=2(I-j)$ which completes the proof.

See also Obstruction classes and Pontrjagin classes (Ex).

2) For any $n$$n$, let $M=X=S^n$$M=X=S^n$. Note that the stable normal bundle $\nu_S^n$$\nu_S^n$ is trivial since the normal bundle of the codimension one embedding $S^n \hookrightarrow \mathbb{R}^{n+1}$$S^n \hookrightarrow \mathbb{R}^{n+1}$ is trivial. Hence any map $f: S^n \to S^n$$f: S^n \to S^n$ extends to a bundle map $(f,\bar{f}): \nu_{S^n} \to \nu_{S^n}$$(f,\bar{f}): \nu_{S^n} \to \nu_{S^n}$ and so it suffices to find degree $d$$d$ maps $f_d: S^n \to S^n$$f_d: S^n \to S^n$ for any $d$$d$.

For $n=1$$n=1$, a degree $d$$d$ map $f_d: S^1 \to S^1$$f_d: S^1 \to S^1$ is given by wrapping the circle $d$$d$ times around itself. By suspending $n-1$$n-1$ times we get a map $\Sigma^{n-1}(f_d): S^n= \Sigma^{n-1}(S^1) \to \Sigma^{n-1}(S^1) = S^n$$\Sigma^{n-1}(f_d): S^n= \Sigma^{n-1}(S^1) \to \Sigma^{n-1}(S^1) = S^n$ which we can show still has degree $d$$d$ by using local degrees as in the previous exercise.

3) We will show that there is a degree one normal map $(f, \bar{f}): F_g \to F_{g'}$$(f, \bar{f}): F_g \to F_{g'}$ if and only if $g \ge g'$$g \ge g'$. Firstly note that, as above, the codimension one embedding $F_g \hookrightarrow \mathbb{R}^3$$F_g \hookrightarrow \mathbb{R}^3$ is trivial and so $\nu_{F_g}$$\nu_{F_g}$ is trivial for all $g$$g$. Hence any map extends to a normal map and it remains to show that there is a degree one map $f: F_g \to F_{g'}$$f: F_g \to F_{g'}$ if and only if $g \ge g'$$g \ge g'$.

If $g \ge g'$$g \ge g'$, then we can get a map $f: F_g \to F_{g'}$$f: F_g \to F_{g'}$ by contracting the complement of the standard $F_{g'} \setminus D^n \subseteq F_g$$F_{g'} \setminus D^n \subseteq F_g$. This is degree one using local degree as in the first exercise.

Let $g < g'$$g < g'$ and suppose there is a map $f^*: H^*(F_{g'}) \to H^*(F_g)$$f^*: H^*(F_{g'}) \to H^*(F_g)$ of cohomology rings sending the generator of $H^2(F_g)$$H^2(F_g)$ to the generator of $H^2(F_{g'})$$H^2(F_{g'})$. Note that $H^*(F_g)$$H^*(F_g)$ is generated by $\alpha_1, \cdots, \alpha_g, \beta_1, \cdots, \beta_g$$\alpha_1, \cdots, \alpha_g, \beta_1, \cdots, \beta_g$ in degree one and $\gamma$$\gamma$ is degree two such that $\gamma= \alpha_i \cup \beta_i$$\gamma= \alpha_i \cup \beta_i$ for $i=1, \cdots, g$$i=1, \cdots, g$ and all other products of the degree one generators zero. This can be lifted from the quotient map $F_g \to \bigvee_{i=1}^g F_1$$F_g \to \bigvee_{i=1}^g F_1$ and the structure of the cohomology ring of $\bigvee_{i=1}^g F_1$$\bigvee_{i=1}^g F_1$ coming from that of $F_1$$F_1$. Since $g < g'$$g < g'$, the map $f^*: \mathbb{Z}^{2g'} = H^1(F_{g'}) \to H^1(F_g)=\mathbb{Z}^{2g}$$f^*: \mathbb{Z}^{2g'} = H^1(F_{g'}) \to H^1(F_g)=\mathbb{Z}^{2g}$ has a non-zero element $\mu= \sum_{i=1}^g (a_i \alpha_i + b_i \beta_i)$$\mu= \sum_{i=1}^g (a_i \alpha_i + b_i \beta_i)$ in its kernel. Suppose, without loss of generality, that $a_j \ne 0$$a_j \ne 0$. Then $\mu \cup \alpha_j = a_j \gamma$$\mu \cup \alpha_j = a_j \gamma$ and so $f^*(\mu \cup \alpha_j) \ne 0 \in H^2(F_g)$$f^*(\mu \cup \alpha_j) \ne 0 \in H^2(F_g)$. However $f^*(\mu \cup \alpha_j) = f^*(\mu) \cup f*(\alpha_j)$$f^*(\mu \cup \alpha_j) = f^*(\mu) \cup f*(\alpha_j)$ and $f^*(\mu)=0$$f^*(\mu)=0$ by hypothesis. This is a contradiction.

This can also be proven by using that a degree one map $f: F_g \to F_{g'}$$f: F_g \to F_{g'}$ induces a surjection $f_*: \pi_1(F_g) \to \pi_1(F_{g'})$$f_*: \pi_1(F_g) \to \pi_1(F_{g'})$ and so a surjection $\Z^{2g} = \pi_1(F_g)^{ab} \to \pi_1(F_{g'})^{ab} = \Z^{2g'}$$\Z^{2g} = \pi_1(F_g)^{ab} \to \pi_1(F_{g'})^{ab} = \Z^{2g'}$ which gives a contradiction as before.