Let $(M_1,g_1)$ and $(M_2,g_2)$ be Riemannian manifolds of the same
dimension, and let $\phi: M_1 \to M_2$ be a smooth map.
We say that $\phi$ is a local
isometry if $g_2 (\phi_* X, \phi_* Y ) = g_1 (X, Y )$
for all $m \in M_1$ and $X, Y \in T_m M_1,$ where $\phi_* : T_m M_1 \to T_{\phi(m)} M_2$ is the
derivative of the map $\phi$ at $m.$
The relation of being locally isometric for Riemannian manifolds is not symmetric, it is of course reflexive: is it transitive?
Best Answer
The relation is transitive.
Let $\phi \colon (M_1,g_1) \to (M_2,g_2)$ and $\psi \colon (M_2,g_2) \to (M_3,g_3)$ be local isometries. The chain rule $(\psi \circ \phi)_\ast = \psi_\ast \circ \phi_\ast$ yields for all $X,Y \in T_{m}M_1$ that $$ \begin{align*} g_{3}((\psi \circ \phi)_\ast X, (\psi \circ \phi)_\ast Y) & = g_{3}(\psi_\ast\phi_\ast X, \psi_\ast\phi_\ast Y) &&\text{chain rule}\\ &= g_{2}(\phi_\ast X, \phi_\ast Y) && \psi \text{ is a local isometry}\\ &= g_{1}(X,Y) && \phi \text{ is a local isometry} \end{align*} $$ so $\psi \circ \phi \colon (M_1,g_1) \to (M_3,g_3)$ is a local isometry.