# Are loops around cross-caps on a sphere homotopic to each other?

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?

• Nitpick: the word is cross-cap. (With or without a hyphen depending on who you ask). – hmakholm left over Monica Nov 12 '16 at 23:24
• @Makholm. Thanks, I will edit and remove the typos. – vinardo Nov 13 '16 at 0:08

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.
• Sorry, I misunderstood what you meant by "linked to a cross cap." The fundamental group of the Klein bottle as a pair of cross caps is $\langle a,b|a^2b^2=1\rangle$ with the two loops being $a$ and $b$, respectively. With a substitution $c=ab$ then the group is $\langle a,c|aca^{-1}c=1\rangle$, which is the usual presentation, and the $b$ loop is $a^{-1}c$. In any case, the two loops are not homotopic. In fact, the abelianization of the second presentation immediately gives $H_1(K)=\mathbb{Z}\oplus\mathbb{Z}_2$ with $a,c$ as generators, and we can see the loops are not homologous. – Kyle Miller Nov 13 '16 at 3:07
• One thing to beware, by the way, is that although the loops in $\mathbb{R}P^2$ are each involutions, in the Klein bottle they are not. – Kyle Miller Nov 13 '16 at 3:08
• Thanks again for your point above. I am not fully convinced. I thought that in the two cross-cup configuration on a sphere, the two generators (relevant to the Klein Bottle fundamental group) where $a$ the loop linked to the first cross-cap (associated to the torsion part of $H_1$) and $b$ the loop linking the two cross-caps. They are not homotopic. I thought that the loop linked to the second cross-cap was homotopic to the first one (the one I called $a$) – vinardo Nov 13 '16 at 3:43