Given $n\geq1$ and points $p\neq q\in\mathbb R^2$, consider the $2(n-1)$-dimensional manifold $\mathcal M_{p,q}^n$ of simple, non-degenerate polylines from $p$ to $q$ having $n$ segments. By "simple" I mean a polyline does not intersect itself. By "non-degenerate" I mean the $n+1$ vertices of a polyline are all different. (On the other hand, it is allowed for consecutive segments to be collinear.)
What is known about $\mathcal M_{p,q}^n$? Is it connected? What if the polylines are restricted to a subset of $\mathbb R^2$, such as the half-space bounded by the line between $p$ and $q$?
I'd love to get a reference that might address my follow-up questions, too!
Let $M_p^n$ denote the space of all simple polylines $L$ with $n$ segments and the initial vertex at $p$. Let $G$ denote the Lie group of Euclidean similarities (compositions of rotations and dilations) fixing $p$. This group acts on $M_p^n$ freely: $gL=L$ implies $g=e$ for every $g\in G$.
For every fixed $q\ne p$, the submanifold $M_{p,q}^n$ forms a cross-section for the action of $G$ on $M_p^n$: $$ M_p^n=\bigcup_{g\in G} g M_{p,q}^n. $$ Hence, the map $$ (g,L)\mapsto gL $$ is a homeomorphism $$ G\times M_{p,q}^n\to M_p^n. $$ Therefore, it suffices to prove connectivity of $M_p^n$. The proof is an induction on $n$. For $n=1$ everything is clear, so assume that connectivity holds for $n-1$ and let's prove it for $n$.
Consider a polyline $L\in M_p^n$; it is a concatenation the concatenation of the edges $$ e_1 * ... * e_{n-1} * e_n. $$ If the last edge $e_n=[p_{n},p_{n+1}]$ is sufficiently short then, with one exception, for every rotation $R$ around $p_n$, the polyline $$ e_1 * ... * e_{n-1} * R(e_n) $$ is still simple. The only exceptional rotation for which this fails is the rotation $R_0$ such that $R_0(e_n)\cap e_{n-1}\ne \{p_n\}$.
Thus, I will do the following. First, scale $e_n$ so that the new edge $e_n'=[p_n,p_{n+1}']$ is sufficiently short. Then apply a rotation $R$ around $p_n$ to the edge $e_n'$ so that the edges $e_{n-1}, R(e_{n}')$ are collinear and intersect only at $p_n$. Now, set $e'_{n-1}:= e_{n-1}\cup R(e_n')$ (a line segment) and form a new polyline $L'$ as the concatenation $$ e_1 * ... * e_{n-2} * e'_{n-1}\in M_{p}^{n-1}. $$ Lastly, use connectivity of $M_{p}^{n-1}$. qed
Remark. An alternative proof can be given by appealing to a much harder theorem, the solution of the "Carpenter's Ruler Problem" (CRP), mentioned by Desire. Namely, choose a positive vector ${\mathbf a}=(a_1,...,a_n)$ and consider the space $$ M^n_p({\mathbf a}) $$ consisting of simple polylines $L\in M^n_p$ whose side-lengths are fixed to be $a_1,...,a_n$. Then the solution of the CRP shows that $M_{p}^{n}({\mathbf a})$ is connected. Fix a point $q\in E^2 -\{p\}$. Given a path $L_t, t\in [0,1]$ in $M^n_p({\mathbf a})$, where the polyline $L_t$ terminates at $q_t\in E^2$, let $g_t\in G$ denote the unique similarity (fixing $p$) such that $g_t(q_t)=q$. By uniqueness, $g_t$ depends continuously on $t$. Hence, we obtain a path $$ g_t L_t\in M^n_{p,q}, t\in [0,1]. $$ Taking $L_t$ such that $L_0=L$ and $L_1$ lies on a straight line, we obtain a continuous path $g_tL_t$ connecting $L$ to a linear polyline from $p$ to $q$. Proving that the space of simple linear polylines from $p$ to $q$ is contractible is an easy exercise as it is naturally homeomorphic to the following open simplex $$ \{{\mathbf a}=(a_1,...,a_n): a_1+...+a_n=1, a_1>0,..., a_n>0\}. $$
Edit. Two more things:
A minor modification of the first proof shows that the space of simple polylines in $M^n_{p,q}(E^k)$ (with fixed end-points) in the Euclidean space $E^k$, is also connected. A bit more work shows that this space of polylines is contractible.
A cute (and not immediate) property of the subset $C^n_{p,q}({\mathbf a})$ of strictly convex simple polylines in $M^n_{p,q}({\mathbf a})$ (in the plane), is that it consists of exactly two contractible components (obtained from each other by reflecting polylines in the straight line $(pq)$). Here a polyline is strictly convex if it is a convex polygonal curve such that all the edges $e_1,...,e_n$ are non-collinear.