# Fundamental Group of the Torus-Figure 8

by sammycaps
Tags: fundamental, torusfigure
 P: 91 So I'm revamping the question I had posted here, after a bit of work. I'm concerned with the homomorphism induced by the inclusion of the Figure 8 into the Torus, and why it is surjective. There seem to be a lot of semi-explanations, but I just wanted to see if the one I thought of makes sense. So, we know that the fundamental group of the Figure 8 is isomorphic to the free product on 2 generators (i.e. of two copies of the integers), and the fundamental group on the torus is isomorphic to the cartesian product of two copies of the integers. So, I don't know if there is a homomorphism j* such that this diagram commutes, for f and g isomorphisms from above, but if there is then this diagram commutes, $\pi$1(Figure 8) $\stackrel{i*}{\longrightarrow}$ $\pi$1(Torus) $\:\:\:\:\:\:\:$$f\downarrow$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$g\downarrow$ $\:\:\:\:\:\:\:\:$$Z$*$Z$$\:\:\:\:$$\stackrel{j*}{\longrightarrow}$ $\:$$Z×Z$ And then we can do something from there. Is that going somewhere, or not at all?
 Sci Advisor HW Helper P: 9,453 would n't you represent a torus as a square with identifications, then push off any loop in the square onto the boundary?
P: 91
 Quote by mathwonk would n't you represent a torus as a square with identifications, then push off any loop in the square onto the boundary?
So, any loop is homotopic to a loop on the boundary? And then any loop on the boundary is a loop of the figure 8? So then would we say the homomorphism induced by inclusion is $i*([a])=[i\circ a]=[a]$, so then this induced homomorphism is surjective?

 Quote by sammycaps So, any loop is homotopic to a loop on the boundary? And then any loop on the boundary is a loop of the figure 8? So then would we say the homomorphism induced by inclusion is $i*([a])=[i\circ a]=[a]$, so then this induced homomorphism is surjective?
you can think of the torus as a figure 8 with a disk attached. The boundary of the disk is attached to the loop aba$^{-1}$b$^{-1}$ on the figure 8. This is what Mathwonk is saying.