Existence of a universal cover of a manifold.

Question

Suppose $M$ is a manifold, topological or smooth etc. As a topological space $M$ is required to be primarily locally homeomorphic to $\Bbb R^n$, with some added things that don't come along with this, like a global Hausdorff condition mentioned here, second countability or paracompactness etc.

Mainly it would seem to rule out certain pathological examples, or to simplify proofs, rather than saying 'Let $M$ be a Hausdorff, Second countable, Manifold$\ldots$' every time.

From algebraic topology the existence of a universal covering space of a topological space $X$, required $X$ to be connected, locally path connected, and semi locally simply connected. In the course I did we said path connected, locally path connected and semi locally simply connected. I believe these are equivalent.

My question is: For a $M$ a manifold, does $M$ satisfy the existence criterion?

Or should I specifically require that $M$ is a connected manifold, and it would seem that locally path connected and semi locally simply connected come from the charts, or local homeomorphisms to $\Bbb R^n$.

Answer 1

You are correct. Because each point in a manifold has a neighborhood homeomorphic to some Euclidean space, any manifold is locally contractible, which implies that is it both locally path connected and locally simply connected. Therefore if we restrict our attention to connected manifolds (which we usually do), we see that all manifolds admit universal covers (and these are also manifolds).