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For a link's complement $X$ there is an infinity cyclic covering $X_\infty$. One can construct $X_\infty$ out of a Seifert surface $F$ for the corresponding link. A method is described in Lickorish's 'An Introduction to Knot Theory' [here].

There are fewer wasy to argue that this construction doesn't depend on the Seifert surface $F$, one is using knowledge of covering spaces. Another one is also described in Lickorish's book, Thm. 7.9, see [here].

My question is about the proof, about the beginning, that is:

A loop $\alpha$ in $X$ lifts to a loop $\hat{\alpha}$ in $X_\infty$ $\Leftrightarrow$ $\hat{\alpha}(0)$ and $\hat{\alpha}(1)$ are in the same copy of $Y(=X\text{-cut-along-}F)\Leftrightarrow \alpha$ intersects $F$ zero times $\Leftrightarrow$ linking number of $\alpha$ with $F$ is zero

I have a problem to understand this equivalences. It would be great if someone could explain what's going on here geometrically? For example, why is $\alpha$ lifted to two different copys of $Y$ if it intersects with $F$?

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Geometrically, the signed intersection number between the surface and a loop, corresponds to the cup product between the corresponding homology classes. Hence the surface defines a homomorphism on the first homology group to $\mathbb Z$, hence on the fundamental group. Also, by covering space theory, a normal covering corresponds to a fundamental group homomorphism, the covering $X_\infty$ being infinite cyclic, a homomorphism to $\mathbb Z$. We get the relation through the following isomorphisms:

$$ H_2(X) \cong H^1(X,\mathbb Z) \cong Hom(H_1(X),\mathbb Z) \cong Hom( \pi_1(X),\mathbb Z) \cong \mathbb Z \text {-coverings of }X $$

It might be important to note that every element in $H_2$ can be represented by a surface. Now the correspondence you know is that an element in the fundamental group lifts to a cover $\hat X$ (we are on the right hand side), if the corresponding homomorphism evaluates it to zero and that happens if the cup product on homology level with the corresponding surface is zero and vice versa.

You can find more details about this relations in this or this or maybe this answer.

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  • $\begingroup$ I really appreciate your answer, thank you for that! But there is a big issue: I am not familiar with any of these homology theories. I know that one can think abour $H_1$ as the abeliaziation of the fundamental group. But anyway, the construction in Lickorish's book doesn't use this, there is just the construction of $X_\infty$ of a Seifert surface and one knows that this is a covering of $X$. So can one understand the proof with more basic knowledge? $\endgroup$ – user444847 May 26 '17 at 13:00
  • $\begingroup$ okay there is an easier explanation that does not use homology, are you familiar with pullbacks? $\endgroup$ – Daniel Valenzuela May 26 '17 at 13:41
  • $\begingroup$ No, but maybe it is pussible to explain with the picture (of the construction of $X_\infty$) in my first link? $\endgroup$ – user444847 May 26 '17 at 13:51
  • $\begingroup$ It is not possible? $\endgroup$ – user444847 May 27 '17 at 14:18

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