Interpret $\mathbb{R}^6 = (\mathbb{R}^2)^3$ as the space of ordered triangles in the real plane (degenerate triangles are included). There is a map $L \colon \mathbb{R}^6 \to \mathbb{R}^3$ sending a triangle to the lengths of its sides. Concretely, $$L(x_1, y_1, x_2, y_2, x_3, y_3) = \left(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}, \sqrt{(x_1-x_3)^2+(y_1-y_3)^2}, \sqrt{(x_2-x_3)^2+(y_2-y_3)^2}\right).$$ Let $B$ be the image, that is, $B = \{(a,b,c) \in \mathbb{R}_{\ge 0}^3 \mid a +b \ge c, a+c \ge b, b+c \ge a\}$. Is there a continuous section $s \colon B \to \mathbb{R}^6$ of $L$, i.e., $L \circ s = \operatorname{id}$?
My thoughts so far: When restricting to non-degenerate triangles, there is such a section. Just fix the first corner at the origin, let the second corner be on the positive $x$-axis and choose the third point of the triangle to be the unique one in the upper half plane to have the right distances to the other ones. One can write down formulas for the points without problems and the resulting map is continuous. A problem arises for degenerate triangles where $(x_1, y_1)$ and $(x_2, y_2)$ have distance zero. The paths $t \mapsto (t, 1, \sqrt{1+t^2})$ and $(t, 1, \sqrt{1-t+t^2})$ both approach $(0,1,1)$ as $t \to 0$. Applying my conjectured $s$ and taking the limit, however, results in different degenerate triangles.
I tried several other candidates for $s$, but there is always a class of degenerate triangles where discontinuities arise. Therefore, I suppose that no such $s$ exists. Unfortunately, I have no idea how to disprove its existence. The only thing I figured out is that we may assume the first two coordinates of $s$ to be constantly zero. Otherwise we may subtract these coordinates from all components of $s$ without changing the distances between the points. I also tried rotating the triangles given by $s(a,b,c)$ so that the second point lies on a fixed ray starting at the origin. This failed as the angle between $(x_2, y_2)$ and the fixed ray is undefined if $(x_2, y_2)$ is zero.
Another idea I had was using the cohomology of the topological spaces involved. But this does not seem helpful as $B$ is contractible.
EDIT — New Idea: If a section $s$ exists, then $L$ is a quotient map. This means $U \subseteq B$ is open only if $L^{-1}(U)$ is open. This statement looks more susceptible to a refutation.