Connected boundary implies $\pi_1(M,\partial M)=0$.

Question

I have two questions: Let $M$ be a compact connected manifold with boundary.

1, If the boundary $\partial\tilde{M} $ of universal covering $\tilde{M}$ is connected, is $\partial M$ connected? How about converse direction, if not, any counterexamples?

2, Does connectedness of boundary $\partial M$ imply $\pi_1(M,\partial M)=0$, if not, any counterexamples?

If $M$ is not necessarily compact, will it be different?

Thanks for your help.

Answer 1

1. Any space that is connected and locally path connected is path connected. Thus, the boundary of the universal covering is path connected, so you can use the cover to push down a path between any two points. I have to think about the converse more.

2. No, take a torus and remove an open disk $D$. Let this space be $X$ and its boundary be $A$ (a circle). Then from the long exact sequence of homotopy groups we get:$\pi_1(A) \rightarrow \pi_1(X) \rightarrow \pi_1(X,A) \rightarrow 0 $. Now we know $\pi_1(A)=\mathbb{Z}$ and $\pi_1(X)=\mathbb{Z}*\mathbb{Z}$ since it is homotopy equivalent to a wedge of circles. Moreover, the image of the inclusion $\pi_1(A)\rightarrow \pi_1(X)$ is not all of $\pi_1(X)$. This means the kernel of the map $\pi_1(X)\rightarrow \pi_1(X,A)$ is not all of $\pi_1(X)$ which means the map is not the zero map, so $\pi_1(X,A) \neq 0$.