I have a possible simple question but I cannot work out the answer by myself.
Suppose we have two cross-caps on a sphere (i.e. two $RP^2$ attached to the sphere by means of a connected sum) and suppose we have two loops each of them linked to each cross-cap. By linked to the cross-cap, I mean that the loop cannot be detached and float free on the sphere because it is not homotopic to the trivial loop (i.e. it is the loop associate to the $Z_2$ component of the $H_1$ homology groups of the $RP^2$).
Are the two loops described above homotopic to each other?
On one hand I know that if I bring the two cross-caps together and I get a Klein Bottle, this has only a "$Z_2-like$" loop in its $H_1$ homology groups and therefore the two loops have to fall in the same path and be homotopic.
On the other hand I see that it is not possible to unlink each loops from its cross-cap and it is not possible to link it to the other cross cap without going twice around any path in it. So I do not see how would be possible to continuously move one loop and bring it to coincide to the other (which is a naïve way of describing homotopy).
Could anyone help me to understand how it works?
Perhaps the confusion is that $X$, two cross caps in a plane, and $K$, a Klein bottle, have different homotopy types. $X$ is homeomorphic $K$ minus a point, which ends up being homotopy equivalent to the wedge sum of two circles. If you keep track of your two paths along through these equivalences, you'll see that each one is a different generator of the circles' fundamental group.
Another way of seeing this is that a cross cap is but a Mobius strip by another name, and two cross caps in a plane is homotopy equivalent to the wedge sum of two Mobius strips, each of which is homotopy equivalent to a circle.
However, if the two loops are in the Klein bottle, imagined as two cross caps joined at an equator circle, then each going around a cross cap just means they are going around that same circle, so they are homotopic.