There is no retraction of the closed disk onto the unit circle

Given: The closed unit disk $D^2$, the unit circle $S^1$, the common basepoint $e_0=(1,0)$, and the inclusion $i:S^1\hookrightarrow D^2$.

If $A$ is a retract of $X$, then the inclusion induces an injective homomorphism on fundamental groups at every basepoint of $A$ (A retract induces an injection on fundamental groups, and a deformation retract induces an isomorphism).

The closed unit disk and unit circle are respectively the Euclidean closed ball ${\bar{B}}_2(0,1)$ and Euclidean sphere $S_2(0,1)$ (Euclidean spheres and closed balls as subspaces of ${\mathbb{R}}^n$).

Every nonempty convex subset of ${\mathbb{R}}^n$ is simply connected (Every nonempty convex subset of ${\mathbb{R}}^n$ is simply connected).

For the geometric unit circle $S^1$ based at $(1,0)$, ${\pi }_1(S^1,(1,0))\cong (\mathbb{Z},+)$ (The trigonometric loops give ${\pi }_1(\{(x,y):x^2+y^2=1\},(1,0))\cong \mathbb{Z}$).

The Euclidean norm is a norm on ${\mathbb{R}}^2$, so it is absolutely homogeneous and satisfies the triangle inequality (Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation).