Let $M$ be a hyperbolic surface with totally geodesic boundary. Taking the double $DM$ of $M$, it is easy to see using Euler characteristic that $DM$ is itself a hyperbolic surface (without boundary). We can obtain a Riemannian metric of constant sectional curvature $-1$ on $DM$ by extending the metric on $M$ in the obvious way. Is this metric smooth? Of course it's smooth except possibly on the boundary of $M$ - the question is whether the metric is smooth there.
This seems to be a fact which is assumed in a lot of sources I've consulted. However, from reading other posts on Stackexchange (this one for instance: https://mathoverflow.net/questions/140386/regularity-of-metric-of-the-double-of-a-riemannian-manifold), I've gathered that a metric obtained in this way is not smooth in general for the double of an arbitrary Riemannian manifold.
The boundary of a hyperbolic surface with totally geodesic boundary is locally modelled on the left half of the upper half-space model $$\{(x,y) \bigm| x \le 0, y>0\} $$ with the usual upper half-space metric $(dx^2 + dy^2)/y^2$. So it suffices to notice that the double of the left half of upper half space is all of upper half space, on which the metric is certainly smooth.