Let $C$ and $C'$ be algebraic (smooth) curves. If morphism $φ:C→C'$ satisfies the condition that $＃φ^{-1}(Q)＝1$ for all $Q∈C'$, why $φ$ is isomorphism ?
If I could prove $φ$ is a degree one morephism, degree 1 morphism between smooth curves is isomorphism.Thus I want to prove $φ$ has degree 1, in other words, corresponding extension of function field has degree 1.
This is not true. Purely inseparable maps of curves give fibers of cardinality one but are not isomorphisms. In characteristic $p$, consider the map $\Bbb P^1\to \Bbb P^1$ by $[x:y]\mapsto [x^p:y^p]$ which on the standard affine opens is of the form $k[t]\to k[t]$ by $t\mapsto t^p$, and this is a bijection.