Exercise 5, § 5

Let $X_0$ be the unit circle of the numerical plane $\mathbf{R}^2$; let $X_1$ be the union of the segments joining the point $(1,0)$ to the points $(3,1/n)$, for $n\in \mathbf{N}^*$; let $X_2$ be the set of points $(x, y)$ of the plane such that $(x-2)^2+y^2= 1$ and $y\leqslant 0$; we put $X = X_0\cup X_1\cup X_2$.

a) Let $Y = X\cup (-X)$. Let $p: Y\rightarrow X$ be the mapping defined as follows: one has $p$(cos($t$), sin($t$)) $=$ (cos(2$t$), sin(2$t$)) for $t\in \mathbf{R}$ and $p(z) =p(-z) =z$ for $z\in X_1\cup X_2$. Prove that $p$ makes $Y$ a covering of degree 2 of $X$.

b) Let $Z_2$ be the set of points of the plane such that $(x-1)^2+y^2= 4$ and $y\leqslant 0$; we put $Z = X_0\cup X_1\cup (-X_1)\cup Z_2\cup (-Z_2)$. Let $q: Z\rightarrow X$ be the mapping defined by $q$(cos($t$), sin($t$)) $=$ (cos(2$t$), sin(2$t$)) for $t\in \mathbf{R},q(z) =q(-z) =z$ for $z\in X_1$ and $q(z) =q(-z) = 1 +\frac{1}{2}z$ for $z\in Z_2$. Prove that $q$ makes $Z$ a covering of degree 2 of $X$.

c) Prove that $q_*\pi_1(Z,(1,0)) =p_*\pi_1(Y,(1,0))$ but that there does not exist a continuous mapping $f: Z\rightarrow Y$ such that $p\circ f=q$.[^1]

[^1]: This example is due to E. C. Zeeman.