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Fundamental Group of the Torus-Figure 8

by sammycaps
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sammycaps
#1
Dec19-12, 08:07 AM
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,

[itex]\pi[/itex]1(Figure 8) [itex]\stackrel{i*}{\longrightarrow}[/itex] [itex]\pi[/itex]1(Torus)
[itex]\:\:\:\:\:\:\:[/itex][itex]f\downarrow[/itex][itex]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[/itex][itex]g\downarrow[/itex]
[itex]\:\:\:\:\:\:\:\:[/itex][itex]Z[/itex]*[itex]Z[/itex][itex]\:\:\:\:[/itex][itex]\stackrel{j*}{\longrightarrow}[/itex] [itex]\:[/itex][itex]ZZ[/itex]

And then we can do something from there.

Is that going somewhere, or not at all?
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mathwonk
#2
Dec19-12, 11:25 AM
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would n't you represent a torus as a square with identifications, then push off any loop in the square onto the boundary?
sammycaps
#3
Dec19-12, 12:47 PM
P: 91
Quote Quote by mathwonk View Post
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 [itex]i*([a])=[i\circ a]=[a][/itex], so then this induced homomorphism is surjective?

lavinia
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Dec19-12, 05:47 PM
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Fundamental Group of the Torus-Figure 8

Quote Quote by sammycaps View Post
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 [itex]i*([a])=[i\circ a]=[a][/itex], 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[itex]^{-1}[/itex]b[itex]^{-1}[/itex] on the figure 8. This is what Mathwonk is saying.

Van kampen's Theorem then gives you the result you are looking for.

BTW: Think about the same ideas for tori with more than one handle.


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