Jtdylktuy

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 $i:Rto T$ 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.

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.