Let $M$ be a smooth Riemannian manifold, and let $p \in M$. Let $U$ be a sufficiently small neighbourhood of $p$, such that $U$ is a normal neighbourhood of each of its points.
We can choose $U$ in such a way that for every $q\in U$, $\exp_q: B_{h_0}^q(0) \to B_{h_0}(q)$ is a diffeomorphism.
Now, we fix a smooth orthonormal frame $F$ of $TM|_U$. For every $q\in U$, we identify $T_qM \sim \mathbb{R}^d$ using $F_q$, so that $\exp_q$ defines normal coordinates on $B_{h_0}(q)$.
Finally, let $\mathcal{R}_{ijkl}(q)$ be the components of the curvature tensor of $M$ at the point $q$, calculated w.r.t the normal coordinates centered around $q$ (using $\exp_q$ and $F_q$ as described above).
I am trying to show the map $q\mapsto \mathcal{R}_{ijkl}(q)$ is smooth, or at least continuous.
I know that the exponential map is smooth when regarded as $\exp:TM|_U \to M$, but I am not sure how to use this here.
Any ideas?
Because $\mathcal R$ is a tensor, its components at $q$ depend only on the basis used at $q$ to compute them. The coordinate basis vectors at $q$ in normal coordinates centered at $q$ are just the values of your given orthonormal frame there. Thus, if $(F_1,\dots,F_n)$ is a given smooth orthonormal frame, your components are $$ \mathcal R_{ijkl}(q) = \mathcal R \big( F_i|_q,F_j|_q, F_k|_q, F_l|_q\big), $$ which depend smoothly on $q$.