0
$\begingroup$

I am trying to show that the identity map

$ id:S_h \vee S_v \rightarrow S_h \vee S_v$ does not lift to L = $(\mathbb{R} \otimes \mathbb{Z}) \cup (\mathbb{Z} \otimes \mathbb{R}) $ via the covering map $q:L \rightarrow S_h \vee S_v$

Here the covering map is defined via a quotient map. $T^2$ the torus can be viewed as a quotient of $\mathbb{R}^2$ under the equivalent relation that sets $(x,y)\sim (x+m,y+n)$ for any $m,n\in\mathbb{Z}$. This maps each unit square in the plane onto the torus. The image of the x-axis under this map is homeomorphic to $S^1$ and denoted by $S_h$ and the image of the y-axis is similarly denoted as $S_v$.

So far, I'm proceeding by contradiction. Suppose the map does lift. Then $id_*(\pi_1(S_h \vee S_v)) \subset p_*(\pi_1(L))$. The fundamental group of a bouquet of circles is $F_2$ (the free group on 2 generators) but what is the fundamental group of the lattice?

I know its not trivial since for example the loop with corners (1,1), (1,2), (2,2), (2,1) is non trivial.

Thank you for any and all help.

*Edited to redefine L

$\endgroup$
0
$\begingroup$

A lift of the $S_h$ component of the identity must lie entirely in some horizontal line. The fundamental group of a line is trivial, and so this loop must be contractible. If there were some lift, then we could write the inclusion of $S^1$ into $S_h$ as a map into this horizontal line followed by the projection. But since the map from $S^1$ into the horizontal line is trivial, the inclusion of $S^1$ into $S_h$ is trivial. This is not the case by van Kampen, so no such lift exists.

Though it isn't your main question, the fundamental group of the lattice is vastly larger than the wedge of two circles, but also somehow smaller. There is a correspondence between subgroups of the fundamental group and covering spaces. It is not too difficult to see that the fundamental group of the lattice has fundamental group that is free on infinitely many generators which is much "larger" than the free group on two generators, but we also have that this infinite free group is a subgroup of the free group on two generators (and it is of infinite index).

This is a way to prove the group theoretic result that a free subgroup of a free group need not be of less rank.

$\endgroup$
  • 1
    $\begingroup$ Can you explain what you mean by trivial as in: "the map from $S^1$ into the horizontal line is trivial." Do you mean that the induced homomorphism on the fundamental group of $S^1$ must map everything to the identity? $\endgroup$ – Math Lady Jan 23 at 2:36
  • $\begingroup$ @MathLady Yes, that is what I meant. If you can write a map as a composition of two maps, one with that property, then your original map induces the zero homomorphism. $\endgroup$ – Connor Malin Jan 23 at 5:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.