Can someone complete the following table?

$\begin{array}{cc} \text{Topology over } \mathbb{R} & \text{Topology over } \mathbb{C} & \text{Algebraic Geometry} \\\\ \hline \mathbb{R} & \mathbb{C} & \mathbb{A}^1 \\\\ \mathbb{R} P^1 \cong S^1 & \mathbb{C} P^1 \cong S^2 & \mathbb{P}^1 \\\\ {}D^1(\mathbb{R}) \cong [0,1] & D^1(\mathbb{C}) \cong [0,1] \times [0,1] & ? \end{array}$

I agree that this question is quite vague. Possible precise interpretations would be:

Is there a scheme $X$ satisfying $X(R) = D^1(R)$ for nice metric rings $R$? Or: Does the endofunctor $(X,x_0,x_1) \mapsto (X \cup_{x_0 \simeq x'_1} X',x_0,x'_1)$ (with a copy $X'$ of $X$) of the category of $k$-schemes (or locally ringed spaces over $k$) equipped with two $k$-rational points have a terminal coalgebra, similar to Freyd's characterization of $[0,1]$?

But I don't want to limit my question to this interpretation. I also would be happy with an algebraic space or stack corresponding to $[0,1]$. Note that $[-1,1]$ "is" the generalized ring and therefore affine scheme $|\mathbb{Z}_{\infty}|$ à la Durov defined as an algebraic submonad of $\mathbb{R}$.

Background: In $\mathbb{A}^1$-homotopy theory the affine line plays the role of the unit interval, right? But when we define homotopies between morphisms of schemes $X \to Y$ in the naive way as morphisms $X \times \mathbb{A}^1 \to Y$, we won't get a transitive relation. The reason is that, although $\mathbb{A}^1$ has two distinguished rational points $0,1$, we don't get $\mathbb{A}^1$ when we glue two copies of $\mathbb{A}^1$, as in the case of the unit interval.