Finite injectivity radius implies compactness Let $(M,g)$ be a complete Riemannian manifold.  Let $\tau:SM\to\mathbb{R}$ denote the cut distance function, where $SM$ is the unit-tangent bundle of $M$.  Let $i_0(p)$ denote the injectivity radius of $M$ at $p$.  
Is it true that:
(1) If $i_0(p)<\infty$ for all $p\in M$, then $M$ is compact?
In Klingenberg's Riemannian Geometry (on p.127) he mentions that if $M$ posses a nontrivial homotopy group $\pi_i(M)$, $i\geq1$, then there exists infinitely many geodesics joining $p$ to $q$ (presumably for any $p$ and $q$).  Is this equivalent to my formulation of the question above?  
Does anyone have a proof of such a result?  I haven't been able to find anything in the literature yet, but this could be because my knowledge of higher homotopy groups is pretty minimal at best, and I'm not exactly sure what I'm looking for.  The one source cited in Klingenberg was a paper by Serre, "Homologie singulière des espaces fibrés", however my french isn't adequate to find what I'm looking for in the paper.
It could also be the stronger statement: 
(2) If $\tau(x,\xi)<\infty$ for all $(x,\xi)\in SM$, then $M$ is compact. 
Is there a proof (for whichever result) that doesn't require higher homotopy groups?
 A: The answer to question (1) is no. For example, if $M$ is the cylinder $\mathbb S^1\times \mathbb R$ with the standard product metric, then the injectivity radius of $M$ at every point is $\pi$, but $M$ is not compact.
However, the answer to (2) is yes. Here are two proofs that don't involve homotopy groups.
First Proof: Suppose $(M,g)$ is complete and $\tau(x,\xi)<\infty$ for all $(x,\xi)\in SM$. Klingenberg's Lemma 2.1.5 shows that the cut distance function $\tau$ is continuous. Let $x\in M$ be arbitrary. The hypothesis implies that $\xi\mapsto \tau(x,\xi)$ is a continuous real-valued function on the set of unit tangent vectors at $x$, which is compact, so this function has an upper bound $C$. Thus the continuous map $\exp_x\colon \overline B_C(0)\to M$ is surjective, so $M$ is the continuous image of a compact set and thus compact.
Second Proof: I'll prove the contrapositive. Suppose $(M,g)$ is complete and noncompact. Klingenberg's Proposition 2.9.3 shows that for every $x\in M$, there is a ray starting at $x$, i.e., a geodesic $c\colon [0,\infty)\to M$ such that $c(0)=x$ and $c$ restricts to a minimizing geodesic segment between any two of its points. If $\xi$ is the unit vector in the direction of $c'(0)$, then $\tau(x,\xi)=\infty$.
I don't understand your question about "Is this equivalent to my formulation of the question above?" Is what equivalent to what, exactly?
