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Why is this sentence true?

For every not nullhomologous loop $f$ without selfintersections on orientable compact surface $M$ there exists a surjection $\phi: H_1(M) \to \mathbb{Z}/8\mathbb{Z}$ such that $f \notin \ker(\phi)$. Where $H_1$ is abelianization map.

I know that we can present $\phi(f)=a_1^{k_1} b_1^{k_2} \cdots a_g^{k_{2g-1}} b_g^{k_{2g}} x_1^{k_{2g+1}} \cdots x_{b-1}^{k_{2g+b-1}}$

where fundamental group of $M$ (with genus $g$ and $b$ boundary components) is: $$ \pi_1(M, *) = \langle a_1, b_1, \ldots , a_g, b_g, x_1, \ldots , x_b \mid [a_1, b_1]\cdots [a_g, b_g]= x_1\cdots x_b \rangle . $$

so I think the point is to show why always there exists at least one $k_i$ such that $k_i \not \equiv 0\pmod{8}$

Thanks for any help.

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    $\begingroup$ Similar to @studiosus answer, there is a self-homeomorphism carrying any nonseparating simple closed curve in a surface to any other one. So you might as well assume that your curve represents a generator. I'm not sure where the $8$ comes from since a much stronger result is true. $\endgroup$ – Cheerful Parsnip Jul 4 '14 at 0:59
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Draw a smooth simple (oriented) loop $a$ on your surface. Since $a$ is not homologycally trivial, it does not separate the surface. Therefore there exists a smooth transversal oriented loop $b$ on the surface which intersects $a$ transversally and in a single point. By choosing orientation on $b$ correctly, you can assume that the oriented intersection number is positive, $+1$. Now, you define a homonorphism $h: H_1(M)\to Z$ by computing oriented intersection number with $b$. The rest you should be able to do yourself. A good reference for such constructions is Guillemin and Pollack "Differential Topology".

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  • $\begingroup$ I guess this shows a much stronger result, that you can map onto $\mathbb Z$. I wonder where the $8$ comes from in the original problem? $\endgroup$ – Cheerful Parsnip Jul 4 '14 at 0:56
  • $\begingroup$ My conjecture, it comes from a professor who wrote this particular problem, maybe he/she is just fond of the number 8, or maybe he/she had some 4-dimensional applications in mind. For this problem, it could have been any number $>1$. $\endgroup$ – Moishe Kohan Jul 4 '14 at 0:59
  • $\begingroup$ Ok, thanks a lot! Now I see this, but could you prove or show me where I can find the proof of sentences "there exists a smooth transversal oriented loop $b$ on the surface which intersects $a$ transversally and in a single point" and "Since $a$ is not homologycally trivial, it does not separate the surface"? $\endgroup$ – Filip Parker Jul 6 '14 at 13:08
  • $\begingroup$ "Since a is not homologycally trivial, it does not separate the surface" why it's true. Is the opposite implication is true: if a is homologycally trivial, it separate the surface? $\endgroup$ – Filip Parker Jul 29 '14 at 12:04
  • $\begingroup$ @FilipParker: Yes, a simple loop is homologically trivial if and only if it separates the surface. If you cannot find an explanation in "Algebraic Topology" by Massey (or Hatcher), or "Differential Topology" by Guillemin and Pollack, let me know. $\endgroup$ – Moishe Kohan Jul 29 '14 at 16:42

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