In perfect elastostatics, the unknown is the displacement $x\mapsto y$, where $x\in\Omega\subset{\mathbb R}^3$ is the reference configuration, and $y\in{\mathbb R}^3$. It obeys to an 2nd-order PDEs. When we rewrite the PDE as a 1st-order system, the unknown becomes the deformation gradient $F(x):=\nabla y$. 


A fundamental constraint is that the matter cannot interpenetrate. In particular, one must have $\det F>0$ everywhere. This is highly non-trivial, in the sense that the set $P$ of $3\times3$ real matrices with positive determinant has a non-trivial homotopy type. Thanks to the polar factorization, $P$ is homeomorphic to the product of ${\bf SPD}\_3$ (a convex cone, something trivial) with ${\bf SO}\_3$ whose fundamental group is ${\mathbb Z}\_2$.


This raises the following questions. Given a domain $\Omega\subset{\mathbb R}^3$, how many connected components are there in $C(\partial\Omega;P)$ ? How can we determine if an $f\in C(\partial\Omega;P)$ can be lifted as an $F\in C(\Omega;P)$ ? How can we determine if an $f\in C(\partial\Omega;P)$ can be lifted as an $F\in C(\Omega;P)$ such that $F$ is homotopic to the identity ?



Elastostatics and homotopy type
Complementing Dmitri's answer: one can see $SO(3)\cong \mathbb{P^3}(\mathbb{R})$ as the 3-skeleton of $K(\mathbb{Z}/2,1)\cong \mathbb{P^\infty}(\mathbb{R})$. Assuming $\Omega$ and its boundary are non-pathological, they will have the homotopy type of 2-dimensional $CW$-complexes. If $X$ is a 2-dimensional $CW$-complex, one can deform any mapping $X\to K(\mathbb{Z}/2,1)$ into a mapping to the 2-skeleton and any homotopy between such mappings into a mapping to the 3-skeleton. So $H^1(X,\mathbb{Z}/2)\cong [X,K(\mathbb{Z}/2,1)]\cong [X,SO(3)]$.


