# The fundamental group of the Lattice - (R x Z) U (Z x R)

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

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.
• 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? – Math Lady Jan 23 at 2:36