Let $M$ be a complete Riemannian manifold of finite volume, with sectional curvatures bounded by some $K>0$ in absolute value. Let $M_{\geq R}$ be the set of points in $M$ with injectivity radius $\geq R$, that is the set of points $x\in M$ for which the restriction of $exp:T_x M\rightarrow M$ to the $R$-ball in $T_x M$ is injective.

Does there exists a closed Riemannian manifold $\hat{M}$ with $\dim{\hat{M}}=\dim{M}$, such that $M$ embeds (Riemannianly) into $\hat{M}$? Does there exists such $\hat{M}$ whose sectional curvatures are at most $C\cdot K$ in absolute value, and whose injectivity radius at least $\epsilon R$, for some $\epsilon,C >0$ which depend only only on $K,R,\dim{M}$?

The approach I tried is to take two copies of the thick part namely $M_{\geq R}\times \{0,1\}$, and connect them via a mapping cylinder (i.e connect them through $\partial M_{\geq R}\times [0,R]$ in the obvious way). Then, I thought to consider 2 open sets $U_j$ containing $M_{\geq R}\times \{j\}$ as well as a bit of the cylinder $(j=0,1)$, and one open set contained in the cylinder, and take a partition of unity w.r.t this cover in order to smooth things out and define a global Riemannian metric. While it seems that $M$ embeds inside $\hat{M}$, I don't see a way to show that the curvature and injectivity radius of $\hat{M}$ didn't change too much.


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