Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fundamental Group of the Torus-Figure 8

  1. Dec 19, 2012 #1
    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]Z[/itex]*[itex]Z[/itex][itex]\:\:\:\:[/itex][itex]\stackrel{j*}{\longrightarrow}[/itex] [itex]\:[/itex][itex]Z×Z[/itex]

    And then we can do something from there.

    Is that going somewhere, or not at all?
    Last edited: Dec 19, 2012
  2. jcsd
  3. Dec 19, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    would n't you represent a torus as a square with identifications, then push off any loop in the square onto the boundary?
  4. Dec 19, 2012 #3
    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?
  5. Dec 19, 2012 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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.
    Last edited: Dec 19, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook