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Suppose we have a degree $d$ map between two closed compact oriented surfaces $f : \Sigma \to \Sigma'$. What does knowing $d$ tell us about the map on $f_\star : H_1(\Sigma, \mathbb{Z}) \to H_1(\Sigma', \mathbb{Z})$?

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  • $\begingroup$ @QiaochuYuan: I think genus $1$ is a special case since there are lots of self-maps from the torus to itself which are higher degree. I doubt there exists a map from a genus 2 surface to itself of degree $\neq0,\pm1$. $\endgroup$ – Cheerful Parsnip Jun 8 '17 at 20:27
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    $\begingroup$ @GrumpyParsnip: There isn't; see this question. $\endgroup$ – Michael Albanese Jun 8 '17 at 21:25
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Here's an example to think about. You can get a map of any degree from a surface of any genus $\Sigma_g$ to a sphere. Take $d$ disjoint disks in $\Sigma_g$ and crush the complement of these disks to a point. Map the disks onto the sphere in an orientation preserving way, with the boundary going to a fixed point on the sphere. This is a degree $d$ map, but the map on $H_1$ is trivial no matter what $d$. So in this case, the map's degree is not related to $H_1$ in any way.

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denote by $I$ and $I'$ the intersection matrices on $\Sigma$ and $\Sigma'$, in standard basis having form $$ \begin{pmatrix} 0 & 1 & 0 & 0 & \ldots \\ -1 & 0 & 0 & 0 &\\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ \vdots & & & & \ddots \end{pmatrix}. $$

then for matrix $A$ of the homomorphism $f_*$ the following condition holds: $$ A\cdot I\cdot A^t=d\cdot I'. $$ here $A^t$ corresponds to the operator $f^*:H^1(\Sigma')\to H^1(\Sigma)$, and the equation follows from the fact that $f^*$ is a ring homomorphism.

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  • $\begingroup$ I tried this formula with two different examples and it didn't work in either one! The simplest one I tried was the map $f$ from a genus $3$ surface to a genus $2$ surface that crushes a subsurface homeomorphic to a torus with open disk removed to a point. Then the matrix $A$ will be a 4 x 6 matrix which is essentially a 4x4 identity matrix with two extra columns of $0$s. The degree of this map is $1$, but $A^t I' A$ is not any multiple of $I$. I also tried looking at genus $3$ as a double cover of genus $2$ but it didn't work in that case either. $\endgroup$ – Cheerful Parsnip Jun 12 '17 at 16:43
  • $\begingroup$ @GrumpyParsnip are you shure that $A^t I' A$ is not equal to $I$? $\endgroup$ – Andrey Ryabichev Jun 13 '17 at 14:50
  • $\begingroup$ Yes. It has zeroes in the lower corner. $\endgroup$ – Cheerful Parsnip Jun 13 '17 at 17:45
  • $\begingroup$ @GrumpyParsnip surely, it is false! the correct statement tells about dual matrix, corresponding the map $H^1\to H^1$ $\endgroup$ – Andrey Ryabichev Jun 14 '17 at 8:46
  • $\begingroup$ I see, that is much more believable and consistent with my genus 0 example. Do you have a reference or sketch of proof? $\endgroup$ – Cheerful Parsnip Jun 14 '17 at 16:06

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