Covering space lifting property Let $p:Y\rightarrow X$ be a covering space and let $x_0 \in X$.
Define a map $\pi_1(X,x_0)\to M(p^{−1}(x_0))$ where the last bit is the set of maps from $p^{−1}(x_0)$ to itself. The map sends $[f]$ to the map $\phi_{[f]}(x)=$ end point of the lift of $f$ starting at $x$.
I need to show $\phi_{[f][g]}=\phi_{[g]}\circ\phi_{[f]}$.
I know that a lift $\overline{f}$ of $f$ is $[f]=[p]∗[\overline{f}]$.
I also need to find what $\phi_{[c]}$ is where $c$ is the constant path at $x_0$.
Finally show $\phi_{[f]}$ is a bijection. This is probably done by showing it is injective and surjective but I'm not sure where to start.
 A: The homomorphism property follows by the unique path-lifting property for covering spaces.  To wit, for $x \in X$ lying above $x_0$, we have $\phi_{[f][g]}(x)$ is the end point of the unique lift of $f * g$ starting at $x$.  But if $\bar{f}_x$ and $\bar{g}_{\bar{f}_x(1)}$ are lifts of $f$ and $g$ starting at $x$ and $\bar{f}_x(1)$ respectively, then $\bar{f}_x * \bar{g}_{\bar{f}_x(1)}$, suitably reparametrized, is a lift of $f * g$ starting at $x$, and its endpoint is precisely the endpoint of $\bar{g}_{\bar{f}_x(1)}$, which by definition is $\phi_{[g]}(\phi_{[f]}(x))$, which proves the homomorphism property.  
If $c$ is the constant path at $x_0$, then given $x \in p^{-1}(x_0)$, the unique lift $\bar{c}$ of $c$ starting at $x$ is the constant path at $x$.  Its endpoint is the point $x$, so $\phi_{[c]}(x) = x$, i.e., $\phi_{[c]}$ is the identity map.  
Finally, in order to show that $\phi_{[f]}$ is a bijection, it suffices to exhibit an inverse function.  In light of the two facts proven above, can you see an inverse?  
