Thanks for the answer, but I'm having trouble understanding it. The manifold we start with is the Euclidean sphere? We now use a small tube and attach a smaller sphere to it? But now you're suggesting to pull back the Euclidean metric through a diffeomorphism and compare the two metrics? Are you suggesting this is a counterexample which shows the question has a negative answer?

I presume what you mean by saying that the injectivity radius can be made arbitrarily small is that for every $\epsilon > 0$ there is a bi-Lipschitz transform such that the injectivity radius of the metric induced by the bi-Lipschitz transform is now less than $\epsilon$? If that's the case, it doesn't exclude it from being $0$.

I mean a noncompact manifold. I don't care whether $\kappa'$ depends on $\kappa$. The $\mathrm{inj}(M,g_i)$ is the infimum of the $\mathrm{inj}(M,g_i,x)$ over each $x$.

Mohan: I'll look for that paper! Thanks.

(Continued...) My question then can be phrased operator theoretically as: Is $d = d_\infty$. But note that unlike the Sobolev spaces which are the completion of $C^\infty \cap L^2$ in the graph norm of $\nabla$, my operator $d$ does not come from such a completion.

(Continued...) Therefore, by a theorem of Kato, these operators are closeable and let the closure be denoted as $d_\infty$ and $\delta_\infty$. Certainly, since we've taken the closure of these operators, $C^\infty_c$ is dense in the graph norm. Now, I can also just take $\delta_c$ and ask for the adjoint of $\delta_c$ which I'll just denote as $d$. That is, we want $d$ to be the operator with largest domain $D(d) \subset $L^2$ such that $(\delta_c u, v) = (u, dv)$ for $v \in D(d)$.

(Continued...) The sobolev space $H^1$ is constructed as the completion of $C^\infty \cap L^2$ in the graph norm of $\nabla$. My setup is slightly different (and please correct me if there is a better way of doing this). Let $d_c$ be the exterior derivative with domain $C^\infty_c$ and $\delta_c$ also have domain $C^\infty_c$. These maps are related by $(d_c u,v) = (u, \delta_c v)$. Defining the $L^2$ spaces as the completion of $C^\infty_c$ as above), we have that $\delta_c$ and $d_c$ are densely defined.