Talk:Bundle structures and lifting problems (Ex)

Exercise 2.1

Assume that there exists a lift $\overline{f}$$<wikitex>; '''Exercise 2.1''' Assume that there exists a lift $\overline{f}$: $X\to\mathrm{hofib}(p)$ of $f$. Then there exist maps $f_1$: $X\to Y$ and $f_2$: $X\times[0,1]\to Z$ such that for all $x\in X$ we have $\overline{f}(x)=(f_1(x),f_2(x,.))$. Since $p\circ\overline{f}=f$ we find that $f_1=f$. Furthermore we have for all $x\in X$: $f_2(x,1)=g(f(x))$ and $f_2(x,0)=z_0$. Thus $f_2$ defines a homotopy from $g\circ f$ to a constant map. Assume that there exists $h$: $X\times[0,1]\to Z$ such that for all $x\in X$ we have $h(x,0)=z_0$ and $h(x,1)=g(f(x))$. Define $\overline{f}$: $X\to\mathrm{hofib}(p)$ by $\overline{f}(x)=(f(x),h(x,.))$. By the definitions of $h$ and of $\mathrm{hofib}(p)$ we find that $\overline{f}$ is well defined and a lift of $f$. '''Exercise 4.1''' * The map $P(K(n+1,\mathbb{Z}/2\mathbb{Z}))\to K(n+1,\mathbb{Z}/2\mathbb{Z})$ is given by $\gamma\mapsto\gamma(1)$. Since $P(K(n+1,\mathbb{Z}/2\mathbb{Z}))$ is contractible, every third term in the long exact sequence in homotopy is zero. Thus we obtain for all $m\geq1$ that $\pi_m(K(n+1,\mathbb{Z}/2\mathbb{Z}))\cong\pi_{m-1}(\Omega K(n+1,\mathbb{Z}/2\mathbb{Z}))$. By the uniqueness of Eilenberg-MacLane spaces the first assertion follows. * It is clear that composition of paths induces a group structure on $[Y,\Omega X]$. Thus the second assertion follows. * Using the long exact sequence of the fibration $O\to EO\to BO$ and that $EO$ is contractible we obtain for all $n\geq1$ that $\pi_n(BO)\cong\pi_{n-1}(O)$. * Using the long exact sequence of the fibration $BSO\to BO\to K(1,\mathbb{Z}/2\mathbb{Z})$ we obtain for all $n\geq2$ that $\pi_n(BSO)\cong\pi_n(BO)$ and $\pi_1(BSO)=0$. Of course $\pi_0(BSO)=0$. * The homotopy fiber of $p$: $BSO\to BO$ can be calculated from the long exact sequence asociated to the fibration $$ \mathrm{hofib}(p)\to BSO\to BO $$ using the previous results. For all $n\geq1$ we obtain $\pi_n(\mathrm{hofib}(p))=0$ and $\pi_0(\mathrm{hofib}(p)))\mathbb{Z}/2\mathbb{Z}$. </wikitex>\overline{f}$: $X\to\mathrm{hofib}(p)$$X\to\mathrm{hofib}(p)$ of $f$$f$. Then there exist maps $f_1$$f_1$: $X\to Y$$X\to Y$ and $f_2$$f_2$: $X\times[0,1]\to Z$$X\times[0,1]\to Z$ such that for all $x\in X$$x\in X$ we have $\overline{f}(x)=(f_1(x),f_2(x,.))$$\overline{f}(x)=(f_1(x),f_2(x,.))$. Since $p\circ\overline{f}=f$$p\circ\overline{f}=f$ we find that $f_1=f$$f_1=f$. Furthermore we have for all $x\in X$$x\in X$: $f_2(x,1)=g(f(x))$$f_2(x,1)=g(f(x))$ and $f_2(x,0)=z_0$$f_2(x,0)=z_0$. Thus $f_2$$f_2$ defines a homotopy from $g\circ f$$g\circ f$ to a constant map.

Assume that there exists $h$$h$: $X\times[0,1]\to Z$$X\times[0,1]\to Z$ such that for all $x\in X$$x\in X$ we have $h(x,0)=z_0$$h(x,0)=z_0$ and $h(x,1)=g(f(x))$$h(x,1)=g(f(x))$. Define $\overline{f}$$\overline{f}$: $X\to\mathrm{hofib}(p)$$X\to\mathrm{hofib}(p)$ by $\overline{f}(x)=(f(x),h(x,.))$$\overline{f}(x)=(f(x),h(x,.))$. By the definitions of $h$$h$ and of $\mathrm{hofib}(p)$$\mathrm{hofib}(p)$ we find that $\overline{f}$$\overline{f}$ is well defined and a lift of $f$$f$.

Exercise 4.1

• The map $P(K(n+1,\mathbb{Z}/2\mathbb{Z}))\to K(n+1,\mathbb{Z}/2\mathbb{Z})$$P(K(n+1,\mathbb{Z}/2\mathbb{Z}))\to K(n+1,\mathbb{Z}/2\mathbb{Z})$ is given by $\gamma\mapsto\gamma(1)$$\gamma\mapsto\gamma(1)$. Since $P(K(n+1,\mathbb{Z}/2\mathbb{Z}))$$P(K(n+1,\mathbb{Z}/2\mathbb{Z}))$ is contractible, every third term in the long exact sequence in homotopy is zero. Thus we obtain for all $m\geq1$$m\geq1$ that $\pi_m(K(n+1,\mathbb{Z}/2\mathbb{Z}))\cong\pi_{m-1}(\Omega K(n+1,\mathbb{Z}/2\mathbb{Z}))$$\pi_m(K(n+1,\mathbb{Z}/2\mathbb{Z}))\cong\pi_{m-1}(\Omega K(n+1,\mathbb{Z}/2\mathbb{Z}))$. By the uniqueness of Eilenberg-MacLane spaces the first assertion follows.
• It is clear that composition of paths induces a group structure on $[Y,\Omega X]$$[Y,\Omega X]$. Thus the second assertion follows.
• Using the long exact sequence of the fibration $O\to EO\to BO$$O\to EO\to BO$ and that $EO$$EO$ is contractible we obtain for all $n\geq1$$n\geq1$ that $\pi_n(BO)\cong\pi_{n-1}(O)$$\pi_n(BO)\cong\pi_{n-1}(O)$.
• Using the long exact sequence of the fibration $BSO\to BO\to K(1,\mathbb{Z}/2\mathbb{Z})$$BSO\to BO\to K(1,\mathbb{Z}/2\mathbb{Z})$ we obtain for all $n\geq2$$n\geq2$ that $\pi_n(BSO)\cong\pi_n(BO)$$\pi_n(BSO)\cong\pi_n(BO)$ and $\pi_1(BSO)=0$$\pi_1(BSO)=0$. Of course $\pi_0(BSO)=0$$\pi_0(BSO)=0$.
• The homotopy fiber of $p$$p$: $BSO\to BO$$BSO\to BO$ can be calculated from the long exact sequence asociated to the fibration

$\displaystyle \mathrm{hofib}(p)\to BSO\to BO$ $$\displaystyle \mathrm{hofib}(p)\to BSO\to BO$$

using the previous results. For all $n\geq1$$n\geq1$ we obtain $\pi_n(\mathrm{hofib}(p))=0$$\pi_n(\mathrm{hofib}(p))=0$ and $\pi_0(\mathrm{hofib}(p)))\mathbb{Z}/2\mathbb{Z}$$\pi_0(\mathrm{hofib}(p)))\mathbb{Z}/2\mathbb{Z}$.