I am facing the following problem of one of the first chapters of a course about Differential Geometry. We have this definition of isometry:
A function $f:S_1 \longrightarrow S_2 $ with $S_1$ and $S_2$ regular surfaces is an isometry if $f$ is a diffeomorphism and $<df_p(v),df_p(w)>=<v,w>$ $\forall p \in S_1, v, \ w \in T_pS_1$.
The problem is: Let $S$ be a connected surface and $f,g:S \longrightarrow S'$ two isometries. Prove that if there is a point $p \in S$ such that $f(p)=g(p)$, $df_p=dg_p$ then $f=g$.
I have tried this but got stuck. Let $q$ be a point of $S$ and let $\alpha$ a regular curve such that $\alpha(0)=p$ and $\alpha(1)=q$. Then $f(\alpha(0))=g(\alpha(0))$ and $df_p(\alpha'(0)) = dg_p(\alpha'(0))$ so that the regular curves $f \circ \alpha$ and $g \circ \alpha$ and its derivatives are equal at $t=0$. I could not prove that $f \circ \alpha = g\circ \alpha$ so that $f(q)=g(q)$.