Simply connected in "An application of second variation to submanifold theory" I read the section 3 of chapter 10 of do Carmo's Riemannian Geometry. In fact, it is reproduce of
Moore, John Douglas, An application of second variation to submanifold theory, Duke Math. J. 42, 191-193 (1975). ZBL0337.53045.
But I can't understand where use the simply connect in the proof.  I have a guess, but clearly something is wrong.
If  $\overline M$ is not simply connect, I have many "holes" on $\overline M$ such that the distance of two holes converge to  $0$. Then, there are $q_i$ such that  $\frac{\partial f}{\partial s}\rightarrow 0$ as $i\rightarrow \infty$. But such manifold is not complete.


Proof in do Carmo's book:







 A: Acturally the mystery lies in the seemingly trivial sentence "Using the formula for the first variation, we see that $\gamma$ is perpendicular to $U$ at q". For a simplyly connected $\overline{M}$, we can apply the Cartan Hardamard theorem to ensure $d(c(s))=L(s)$, where $c(s)$ is a curve in $U$ with $c(0)=q$ and $L(s)$ represents the arc lenth of geodesics given by $t\mapsto \gamma_s(t)=\mathrm{exp}_{\overline{p}}\frac{t}{l}(\mathrm{exp}_{\overline{p}}^{-1}(c(s)))$. Then, indeed, since $c(0)=q$ is a critical point of $d$, and by first variation $L'(0)=\frac{d}{ds}\Big{|}_{s=0}d(c(s))=0$ we arrive at $\gamma\perp U$. However, if $\overline{M}$ is not simply connnected, $d(c(s))=L(s)$ may be wrong! The simplest counter example is as follows: Consider $\overline{M}=S^1\times\mathbb{R}$. $S^1\times\{1\}\subset \overline{M}$ is an embedded submanifold. Let $\overline{p}=(e^{\mathrm{i}\theta},0)$, then $q=(e^{-\mathrm{i}\theta},1)\in S^1\times\{1\}$ is the farest point in $S^1\times\{1\}$ from $\overline{p}$. Now consider a curve $c(s)=(e^{-\mathrm{i}(\theta+s)},1)$ passing through $q$. But clearly $d(c(s))\neq L(s)$ since $c(s)$ passes through the cut locus of $\overline{p}$ and hence the minimizing geodesic is no longer $\gamma_s(t)$! And it's not hard to see in this example the minimizing geodesic connecting $\overline{p}$ and $q$ is not perpendicular to $S^1\times\{1\}$.
