A quite direct argument is to solve exercise 4.4.17 in page 252 of Thurston's book 3-D Geom & Topo - the published one in 1997. When doing this exercise, one may want to notice two things: 1. On $S^2$, each of the $5$ angles of a spherical pentagon face of a dodecahedron is $120$ degrees, in contrast to $108$ degrees for a pentagon on plane - this is useful when you try to create a local coordinate system on the circle bundle $UTS^2$; 2. The "fundamental domain" Thurston created here does not look like a dodecahedron in the metric sense - it look like a cylinder over a pentagon base; however other "fundamental domains" will mark a line (with quite a slope!) on each of the five side faces and therefore each side face splits into two "pentagons"... With these in mind, exercise 4.4.17 should be very pleasant to work out.

A quite direct argument is to solve exercise 4.4.17 in page 252 of Thurston's book 3-D Geom & Topo - the published one in 1997; in this exercise Thurston considered the action of the icosahedral group on the unit circle bundle $UTS^2=RP^3$; this is a free action, and the point is to describe a fundamental domain and show that the pattern of gluing of its boundary is exactly the same as in the Poincare dodecahedral space. When doing this exercise, one may want to notice two things: 1. On $S^2$, each of the $5$ angles of a spherical pentagon face of a dodecahedron is $120$ degrees, in contrast to $108$ degrees for a pentagon on plane - this is useful when you try to create a local coordinate system on the circle bundle $UTS^2$; 2. The fundamental domain Thurston created here does not look like a dodecahedron in the metric sense - it look like a cylinder over a pentagon base; however other fundamental domains will mark a line, with quite a slope, on each of the five side faces and therefore each side face splits into two "pentagons"... With these in mind, exercise 4.4.17 should be very pleasant to work out.

A quite straightforward way of seeing that the fundamental group is binary icosahedral is to solve exercise 4.4.17 in page 252 of Thurston's book 3-D Geom & Topo (the published one in 1997). When doing this exercise, one should notice two things: 1. on $S^2$, each of the $5$ angles of a pentagon face of a dodecahedron is $120$ degrees, in contrast to  $108$ degrees for a pentagon on plane - this is useful when you try to create a coordinate system on the circle bundle $UTS^2$; 2. the "fundamental domain" Thurston created here does not look like a dodecahedron in the metric sense - it has only $5$ side faces, a pentagon top face and a pentagon bottom face; however other "fundamental domains" will mark a line on each of the side faces and therefore each side face splits into two "pentagons"... With these in mind, exercise 4.4.17 will be very pleasant to work out.

A quite direct argument is to solve exercise 4.4.17 in page 252 of Thurston's book 3-D Geom & Topo - the published one in 1997. When doing this exercise, one may want to notice two things: 1. On $S^2$, each of the $5$ angles of a spherical pentagon face of a dodecahedron is $120$ degrees, in contrast to  $108$ degrees for a pentagon on plane - this is useful when you try to create a local coordinate system on the circle bundle $UTS^2$; 2. The "fundamental domain" Thurston created here does not look like a dodecahedron in the metric sense - it look like a cylinder over a pentagon base; however other "fundamental domains" will mark a line (with quite a slope!) on each of the five side faces and therefore each side face splits into two "pentagons"... With these in mind, exercise 4.4.17 should be very pleasant to work out.