Fundamental groups of the configuration spaces of all triangles and right triangles This is a question from a past comprehensive exam:
Consider triangles in the plane, with vertices given by non-colinear points as usual. The space $T$ of all plane triangles can be given a natural quotient topology: $T$ is the quotient of an open subset of $\mathbb{R}^6=(\mathbb{R}^2)^3$ by the action of the symmetric group $S_3$ permuting the vertices of a triangle.
Let $R$ be the space of all right triangles in the plane. Let $i:R\to T$ be the inclusion map. Is the map $i_*:\pi_1(R)\to \pi_1(T)$ surjective?
I do not know the answer and I only have a very vague picture of $T$ and $R$ as some fiber bundles. I don't really know where to start when faced with such problems. Any hint will be appreciated.
 A: First, consider the map that sends a planar triangle to its circumcircle.  A circle in the plane is determined by its center in $\mathbb{R}^2$ and its radius in $\mathbb{R}_{>0}$, so the space of circles is homeomorphic to $\mathbb{R}^2 \times \mathbb{R}_{>0}$, which in any case is contractible.  This define maps of fiber bundles 
\begin{array}{ccccc}
F_R & \rightarrow & R & \rightarrow & \mathbb{R}^2 \times \mathbb{R}_{>0} \simeq * \\
\downarrow & & \downarrow & & \downarrow \\
F_T & \rightarrow & T & \rightarrow & \mathbb{R}^2 \times \mathbb{R}_{>0} \simeq *
\end{array}
We are reduced to studying the map of fibers $F_R \to F_T$.  Let us identify these spaces more explicitly.  First, $F_T = \mathrm{UConf}_3(S^1)$ is just the unordered configuration space of three distinct points on a circle.  On the other hand, the projection of a configuration in $F_R$ to the vertex with the right angle determines a fiber bundle $$S^1 \setminus \{*\} \to F_R \to S^1,$$ which means $F_R \xrightarrow{\sim} S^1$ as $S^1 \setminus \{*\}$ is contractible.  Note that a generator of $\pi_1 F_R \cong \mathbb{Z}$ is represented by a full revolution around the circle.   
Now, we have $$\mathrm{UConf}_3(S^1) = \mathrm{Conf}_3(S^1)/\Sigma_3 \simeq (S^1 \sqcup S^1)/\Sigma_3 \cong S^1 / C_3.$$  Let me explain this a little.  First, $\mathrm{Conf}_3(S^1)$ has two path components corresponding to the clockwise and counterclockwise cyclic orderings of the points, and each component fibers over the circle with fiber a contractible simplex, so $\mathrm{Conf}_3(S^1) \simeq S^1 \sqcup S^1$. There is a transposition in $\Sigma_3$ that identifies the two cyclic orderings, and thus we are left with $\mathrm{UConf}_3(S^1) \simeq S^1 / C_3$.  Here, a generator of  $\pi_1 \mathrm{UConf}_3(S^1) \cong \mathbb{Z}$ is represented by a partial revolution of the circle.  
Combining these observations, we have
\begin{array}{ccc}
\pi_1 F_R & \rightarrow & \pi_1 F_T \\
\downarrow & & \downarrow \\
\mathbb{Z} \cong \pi_1 S^1 & \xrightarrow{3} & \pi_1 (S^1/C_3) \cong \mathbb{Z} 
\end{array}
So it seems that the map induced on $\pi_1$ by the inclusion $R \to T$ is not surjective.  
