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I think the map $F$ is injective, but not surjective.

There is a unique map of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$, up to the action of $PGL_2(\mathbb{R})$. If we parameterize the slits as $[a_0,a_1],[a_2,a_3],[a_4,a_5]$, where $a_0 < a_1 < a_2 < a_3 < a_4 < a_5$ in $\mathbb{R}$, then the configuration is invariant under complex conjugation, which induces a hyperbolic isometry of the pants. So the pant seams are the intervals $[a_1,a_2], [a_3,a_4], [a_5,a_0]$ taken in cyclic order on the circle $\mathbb{RP}^1$. The rings $R_i$ are the complements of the slits $[a_i,a_{i+1}]\cup [a_{i+3},a_{i+4}]$, indices taken $(\mod 6)$ (where $R_i \cong R_{i+3}$). The points are uniquely determined by the modulus of $R_i$, up to the action of $PSL(2,\mathbb{R})$, or by the cross-ratio $[a_i,a_{i+1};a_{i+3},a_{i+4}]$.

Therefore the modulus of each ring determines uniquely another ring $U_i$ which is the complement of the slits $[a_{i+1},a_{i+3}]\cup [a_{i+4},a_i]$. Then $R_{i+1} \supset U_i$, $R_{i+2}\supset U_i$. So the moduli of the rings $R_{i+1}$ and $R_{i+2}$ bound the modulus of the ring $U_i$, which uniquely determines the modulus of $R_i$. So this shows one cannot achieve all possible triples of moduli, so the map $F$ is not surjective.

I think the map $F$ is injective, but not surjective.

There is a unique conformal map of the interior of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$, up to the action of $PGL_2(\mathbb{R})$. If we parameterize the slits as $[a_0,a_1],[a_2,a_3],[a_4,a_5]$, where $a_0 < a_1 < a_2 < a_3 < a_4 < a_5$ in $\mathbb{R}$, then the configuration is invariant under complex conjugation, which induces a hyperbolic isometry of the pants. So the pant seams are the intervals $[a_1,a_2], [a_3,a_4], [a_5,a_0]$ taken in cyclic order on the circle $\mathbb{RP}^1$. In fact, one may think of the map as "sewing" the cuffs of the pants along the two intervals connecting the seams. The rings $R_i$ are the complements of the slits $[a_i,a_{i+1}]\cup [a_{i+3},a_{i+4}]$, indices taken $(\mod 6)$ (where $R_i \cong R_{i+3}$). The points are uniquely determined by the modulus of $R_i$, up to the action of $PSL(2,\mathbb{R})$, or by the cross-ratio $[a_i,a_{i+1};a_{i+3},a_{i+4}]$.

Therefore the modulus of each ring determines uniquely another ring $U_i$ which is the complement of the slits $[a_{i+1},a_{i+3}]\cup [a_{i+4},a_i]$ (where the intervals are determined by the cyclic order and orientation of $\mathbb{RP}^1$). Then $R_{i+1} \supset U_i$, $R_{i+2}\supset U_i$, as homotopy equivalences. So the moduli of the rings $R_{i+1}$ and $R_{i+2}$ bound the modulus of the ring $U_i$, by the monotonicity of moduli of annuli, which uniquely determines the modulus of $R_i$. So this shows one cannot achieve all possible triples of moduli, so the map $F$ is not surjective.

I think the map $F$ is injective, but not surjective.

There is a unique map of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$. If we parameterize the slits as $[a_0,a_1],[a_2,a_3],[a_4,a_5]$, then the configuration is invariant under complex conjugation, which is a hyperbolic isometry. So the seams are the intervals $[a_1,a_2], [a_3,a_4], [a_5,a_0]$ taken in cyclic order on the circle $\mathbb{RP}^1$. The rings $R_i$ are the complements of the slits $[a_i,a_{i+1}]\cup [a_{i+3},a_{i+4}]$, indices taken $(\mod 6)$. The points are uniquely determined by the modulus of $R_i$, up to the action of $PSL(2,\mathbb{R})$, or by the cross-ratio $[a_i,a_{i+1};a_{i+3},a_{i+4}]$.

Therefore the modulus of each ring determines uniquely another ring $U_i$ which is the complement of the slits $[a_{i+1},a_{i+3}]\cup [a_{i+4},a_i]$. Then $R_{i+1} \supset U_i$, $R_{i+2}\supset U_i$. So the moduli of the rings $R_{i+1}$ and $R_{i+2}$ bound the modulus of the ring $U_i$, which is uniquely determined by the modulus of $R_i$. So this shows one cannot achieve all possible triples of moduli, so the map $F$ is not surjective.

I think the map $F$ is injective, but not surjective.

There is a unique map of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$, up to the action of $PGL_2(\mathbb{R})$. If we parameterize the slits as $[a_0,a_1],[a_2,a_3],[a_4,a_5]$, where $a_0 < a_1 < a_2 < a_3 < a_4 < a_5$ in $\mathbb{R}$, then the configuration is invariant under complex conjugation, which induces a hyperbolic isometry of the pants. So the pant seams are the intervals $[a_1,a_2], [a_3,a_4], [a_5,a_0]$ taken in cyclic order on the circle $\mathbb{RP}^1$. The rings $R_i$ are the complements of the slits $[a_i,a_{i+1}]\cup [a_{i+3},a_{i+4}]$, indices taken $(\mod 6)$ (where $R_i \cong R_{i+3}$). The points are uniquely determined by the modulus of $R_i$, up to the action of $PSL(2,\mathbb{R})$, or by the cross-ratio $[a_i,a_{i+1};a_{i+3},a_{i+4}]$.

Therefore the modulus of each ring determines uniquely another ring $U_i$ which is the complement of the slits $[a_{i+1},a_{i+3}]\cup [a_{i+4},a_i]$. Then $R_{i+1} \supset U_i$, $R_{i+2}\supset U_i$. So the moduli of the rings $R_{i+1}$ and $R_{i+2}$ bound the modulus of the ring $U_i$, which uniquely determines the modulus of $R_i$. So this shows one cannot achieve all possible triples of moduli, so the map $F$ is not surjective.

I think the map $F$ is injective, but not surjective.

There is a unique map of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$. If we parameterize the slits as $[a_0,a_1],[a_2,a_3],[a_4,a_5]$, then the configuration is invariant under complex conjugation, which is a hyperbolic isometry. So the seams are the intervals $[a_1,a_2], [a_3,a_4], [a_5,a_0]$ taken in cyclic order on the circle $\mathbb{RP}^1$. The rings $R_i$ are the complements of the slits $[a_i,a_{i+1}]\cup [a_{i+3},a_{i+4}]$, indices taken $(\mod 6)$. The points are uniquely determined by the modulus of $R_i$, up to the action of $PSL(2,\mathbb{R})$, or by the cross-ratio $[a_i,a_{i+1};a_{i+3},a_{i+4}]$. Therefore the modulus of each ring determines uniquely another ring $\hat{R}_i$ which is the complement of the slits $[a_{i+1},a_{i+3}]\cup [a_{i+4},a_i]$. Then $R_{i+1} \subset \hat{R}_i$, $R_{i+2}\subset \hat{R}_i$. So the moduli of the rings $R_{i+1}$ and $R_{i+2}$ are bounded by the modulus of the ring $\hat{R}_i$, which is uniquely determined by the modulus of $R_i$. So this shows one cannot achieve all possible triples of moduli, so the map $F$ is not surjective.

I think the map $F$ is injective, but not surjective.

There is a unique map of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$. If we parameterize the slits as $[a_0,a_1],[a_2,a_3],[a_4,a_5]$, then the configuration is invariant under complex conjugation, which is a hyperbolic isometry. So the seams are the intervals $[a_1,a_2], [a_3,a_4], [a_5,a_0]$ taken in cyclic order on the circle $\mathbb{RP}^1$. The rings $R_i$ are the complements of the slits $[a_i,a_{i+1}]\cup [a_{i+3},a_{i+4}]$, indices taken $(\mod 6)$. The points are uniquely determined by the modulus of $R_i$, up to the action of $PSL(2,\mathbb{R})$, or by the cross-ratio $[a_i,a_{i+1};a_{i+3},a_{i+4}]$. 

Therefore the modulus of each ring determines uniquely another ring $U_i$ which is the complement of the slits $[a_{i+1},a_{i+3}]\cup [a_{i+4},a_i]$. Then $R_{i+1} \supset U_i$, $R_{i+2}\supset U_i$. So the moduli of the rings $R_{i+1}$ and $R_{i+2}$ bound the modulus of the ring $U_i$, which is uniquely determined by the modulus of $R_i$. So this shows one cannot achieve all possible triples of moduli, so the map $F$ is not surjective.