Compute the fundamental group, using any technique you like, of \(\RP^2 \#\RP^2 \#\RP^2\).
Solution
Let \[ N_3=\RP^2\#\RP^2\#\RP^2. \]
1. The surface \(N_3\) has a CW structure with one \(0\)-cell, three \(1\)-cells \(a,b,c\), and one \(2\)-cell whose attaching word is\[ a^2b^2c^2. \]
Remove the interior of a disk from \(N_3\). The resulting nonorientable surface \(N_{3,1}\) is a disk with three twisted bands attached, one for each projective-plane summand. Collapse the disk and the transverse directions in the three bands onto their cores. This gives a deformation retraction \[ N_{3,1}\simeq S^1_a\vee S^1_b\vee S^1_c. \]
When the boundary of \(N_{3,1}\) is traversed once, the core of each twisted band is traversed twice in the same direction: the twist carries one edge of the band to the other edge with the same orientation, so the two traversals contribute \(a^2\), and similarly for the other two bands. Ordering the bands as \(a,b,c\), the boundary loop therefore represents \[ a^2b^2c^2 \] in \(\pi_1(N_{3,1})\). Gluing back the removed disk attaches one \(2\)-cell along precisely this boundary loop. Hence \(N_3\) has the asserted CW structure.
2. Therefore\[ \boxed{ \pi_1(\RP^2\#\RP^2\#\RP^2) \cong \left\langle a,b,c\ \middle|\ a^2b^2c^2=1\right\rangle . } \]
The \(1\)-skeleton in 1 is a wedge of three circles, so its fundamental group is the free group \[ F(a,b,c). \] Attaching a \(2\)-cell along the loop represented by \(a^2b^2c^2\) quotients this free group by the normal closure of that word. The Seifert–van Kampen theorem therefore gives \[ \pi_1(N_3) \cong F(a,b,c)/\!\left\langle\!\left\langle a^2b^2c^2\right\rangle\!\right\rangle, \] which is exactly the displayed presentation.