Let p: E--->B be a covering map with E connected, and p(e0)=b0. Show that the homomorphism p* : π1(E; e0)--->π1(B; b0) is one to one.

Question

Let $p: E \to B$ be a covering map with $E$ connected, and $p(e_0)=b_0$. Show that the homomorphism $p* : \pi_1(E; e_0) \to \pi_1(B; b_0)$ is one to one.

Should I need to show p is a homeomorphism?

Answer 1

Let $[\alpha]$ and $[\beta]$ be two path classes in $E$, i.e. two elements in $\pi_1(E,e_0)$, and suppose that $f_\alpha$ and $g_\beta$ are loops in $[\alpha]$ an $[\beta]$ respectively.

If $p_*[\alpha] = p_*[\beta]$ then $p \circ f_\alpha \simeq p\circ f_\beta$, by definition of $p_*$.

A standard result on homotopies then gives that $f_\alpha \simeq f_\beta$ and hence $[\alpha] = [\beta]$.