Talk:Bundle structures and lifting problems (Ex)

Revision as of 19:12, 2 April 2012

Exercise 2.1

Assume that there exists a lift $<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 $f$. </wikitex>\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 $f$.