Determine the fundamental class of $D^2$, where $D$ is a disk.

Question

I have the following definitions.

The fundamental group of X with basepoint $x_0$ is given by:

$\pi_1(X,x_0)$ := {homotopic equivalence classes of loops in X with basepoint $x_0$}.

The open disk of center (a,b) and radius R is given by the formula:

$D = \big{\{}(x,y)\in\mathbb{R}^2:(x-a)^2 + (y-b)^2 < R^2\big{\}}$.

Now, I'm asked to determine $\pi_1(D^2)$. Is there no basepoint given because it's not relevant which basepoint we use? (Give that the basepoint does lie in the disk we're looking at). If I were to look at $\pi_1(D)$, I'd say that $\pi_1(D)$ equals the trivial set, since every pair of loops within D is homotopic. But when it concerns $\pi_1(D^2)$ I'm not so sure.

Q: What is $\pi_1(D^2)$?

Answer 1

1) One should talk about "fundamental group" as "fundamental class" has a different meaning in algebraic topology.

2) You are right for $D$, you can show by hand that $\pi_1(D^2) = 0$ (try to find an homotopy between any path and the constant path $(O,O)$ where $O$ is the center of the disk) or noticing that $\pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y)$ for any topological spaces $X,Y$.

Answer 2

$D^2$ is convex, hence its fundamental group is trivial.