Talk:Bundle structures and lifting problems (Ex)

Exercise 2.1

Assume that there exists a lift $\overline{f}$$<wikitex>; '''Exercise 2.1''' Assume that there exists a lift $\overline{f}$: $X\to\mathrm{hofib}(p)$ of $f$. Then there exist maps $f_1$: $X\to Y$ and $f_2$: $X\times[0,1]\to Z$ such that for all $x\in X$ we have $\overline{f}(x)=(f_1(x),f_2(x,.))$. Since $p\circ\overline{f}=f$ we find that $f_1=f$. Furthermore we have for all $x\in X$: $f_2(x,1)=g(f(x))$ and $f_2(x,0)=z_0$. Thus $f_2$ defines a homotopy from $g\circ f$ to a constant map. </wikitex>\overline{f}$: $X\to\mathrm{hofib}(p)$$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.