Path-connected iff connected on extended complex plane

Question

The following is true for an open set $D \subset \mathbb{C}$:

$D$ is connected $\iff$ $D$ is path-connected.

The latter means there is a continuous function $\gamma: [0,1] \to D$ for $z_0, z_1 \in D$ so we have $\gamma(0)=z_0$ and $\gamma(1)=z_1$.

Now, I want to consider the statement on the extended complex plane $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$.

I have read that $\hat{\mathbb{C}}$ is a connected topological manifold which is locally path-connected and thus path-connected.

I must admit that I don't feel comfortable with this statement, probably because I'm not that familiar with these topological terms. Maybe one of you can clarify it or even show the statement for $\hat{\mathbb{C}}$ in a more direct way.

If $z_0, z_1 \in D \subset \hat{\mathbb{C}}, z_1=\infty$, then $\gamma(t)=(1-t) z_0 + \frac{t}{1-t}, t\in[0,1]$ is a continuous function so that $z_0$ and $z_1$ are path-connected on $D$?

Help is appreciated!

Answer 1

A compact space that is connected and locally path-connected is path-connected. This is not a difficult exercise. (cover X by finitely many path-connected sets).

Alternatively, just prove directly that the Riemann sphere is path-connected. Indeed, you can connect any two points of the complex plane by a line segment, and any finite point with infinity by a half line ...