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Let $X_2$ and $X_3$ denote closed, oriented surfaces of genus $2$ and $3$ respectively.

  • What is the homology of $X_2$ and $X_3$?
  • What is the homology of the product $X_2 \times X_3$?
  • What are the maps on homology induced by projections onto the two factors?
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2 Answers 2

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$X_i$ be the $i(=2,3)$-genus closed orientable surface. So to compute $H_1$ we need to see the structure of fundamental group. As we know that a $g-$genus orieted surface has the cell structure with one $0-cell$, $2g$ $1-cells$, and one $2-cell$. The $1-skeleton$ is a wedge sum of $2g$-circles and the two cell is attached along the loop $[a_1,b_1]\cdots [a_d,b_g]$. There fore $H_1(X_i)=AB(\pi_1(X_i))= \mathbb Z^{2i}$. Since they are connected and oriented, so $H_0(X_i)=H_2(X_i)=\mathbb Z$.

No use Kunneth formula for homology to get 2nd answer. Observe that all homology geoups are free. So $H_k(X_2\times X_3)= \sum_{i+j=k} H_i(X_2)\otimes H_j(X_3)$.(just compute it)

And from this formula we can now see that projection map $p_*:H_i(X_2\times X_3)\to H_i(X_j)$ sends the $i-th$ homology of $X_j$ onto its cannonical image and rest of the generators to zero.

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Since it is most likely a homework problem, I can only sketch the ideas and the necessary keywords.

  • $X_g$ is a connected sum $T\#\cdots\# T$ of $g$ copies of the torus $T$. You can reduce the calculations to the homology of the (punctured) torus and homology of connected sums. For the latter, the Mayer-Vietoris long exact sequence may be of some help.

  • Alternatively, note that you know $H_0 (X_g)$ and $H_2 (X_g)$ (what these groups are and why?), and the only missing group is $H_1 (X_g)$, which is the abelianization of the fundamental group $\pi_1 (X_g)$. You can describe $\pi_1 (X_g)$ in terms of generators and relations, and then compute the abelianization.

  • If you like, you can also consider some simple CW-complexes for $X_g$ and calculate everything using celular homology.

  • The tool to express the homology of $X\times Y$ in terms of homologies of $X$ and $Y$ is the Künneth formula. It is deduced from the Eilenberg-Zilber theorem which says that $C_\bullet (X\times Y) \cong C_\bullet (X) \otimes C_\bullet (Y)$.

For details, see any textbook on algebraic topology.

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