# Embedding a Riemannian manifold with boundary in a closed manifold

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.