# Computing the cohomology ring of the orientable genus $g$ surface by considering a quotient map.

I'm attempting to do Hatcher Chapter 3.2, Exercise 1.

Let $M_g$ be the genus $g$ orientable surface, and let $q$ be the quotient map from $M_g$ to the wedge product of $g$ tori obtained by collapsing a subspace of $M_g$ homeomorphic to a $g$-times punctured sphere to a point.

We know that $H^*(T^2) = \wedge[\alpha,\beta]$, so $H^*(\vee_i T^2) = \oplus_{i = 1}^g \wedge[\alpha_i,\beta_i]$. The map $q$ induces a ring homomorphism $q^*: H^*(\vee_i T^2) \to H^*(M_g)$. Choosing appropriate representatives for the cohomology classes $\alpha_i$ and $\beta_i$ (which I do not denote any differently), we can see that $q^*(\alpha_i)$ and $q^*(\beta_i)$ are represented by $\alpha_i \circ q$ and $\beta_i \circ q$ respectively, and the ring homomorphism preserves the exterior algebra relations.

My only problem is determining if there are other relations in $H^*(M_g)$ because I don't believe $q^*$ is an isomorphism. Is there some way to determine these without directly computing the cup product structure of $M_g$ via a gluing diagram?

The comments on the previous answer express a desire for additional details, so I'll add some here for any future students looking at this question. Robin nicely explains how the only interesting multiplicative structure inside $$H^*(M_g)$$ comes from the cup product $$H^1(M_g) \times H^1(M_g) \to H^2(M_g)$$ so we focus on that. Fix the following notation: let $$X := \bigvee_{i = 1}^g T_i$$ with each $$T_i$$ a torus, let $$A$$ denote the subspace of $$M_g$$ equal to a $$g$$-punctured sphere, let $$i : A \to M_g$$ denote inclusion, and let $$q: M_g \to M_g/A = X$$ denote the quotient map. As Robin mentions, the strategy is use the fact (Hatcher proposition 3.10) that the square $$\require{AMScd} \begin{CD} H^1(X) \times H^1(X) @>{\cup}>> H^2(X) \\ @V{q^* \times q^*} VV @V{q^*}VV \\ H^1(M_g) \times H_1(M_g) @>{\cup}>> H^2(M_g) \end{CD}$$ commutes to get at the cup product structure on $$M_g$$ from (i) the (known) cup product structure on $$X$$ and (ii) the action of $$q$$ on cohomology. As (i) is addressed in the question and in Robin's answer, we focus on the details of (ii).

Disclaimer: I am still very much a student of algebraic topology, so I welcome any comments for improving my argument!

We consider a piece $$\cdots \rightarrow H_1(A) \xrightarrow{i_*} H_1(M_g) \xrightarrow{q_*} H_1(X) \xrightarrow{\delta} H_0(A) \xrightarrow{i_*} H_0(M_g) \to \cdots$$ of the long exact sequence of homology induced by the pair $$(M_g, A)$$. Regard $$H_1$$ as the abelianisation of $$\pi_1$$ to see that the map $$i_* : H_1(A) \to H_1(M_g)$$ is zero (because loops in the subspace $$A$$ lie in the commutator subgroup of $$\pi_1(M_g)$$). Note also that $$i_* : H_0(A) \to H_0(M_g)$$ injects (by the definition of the zero-th homology) so that $$\delta = 0$$. Thus, $$q_*$$ is encased by zero maps and so induces an isomorphism on first homology. By the universal coefficient theorem for cohomology (noticing that the Ext terms vanish here), it follows that $$q^*$$ induces an isomorphism on first cohomology.

Understanding $$q^*$$ on second cohomology requires a bit more topology and in this case we'll think about $$H_2$$ in terms of simplicial homology. Let $$q_i : X \to T_i$$ denote the quotient map which collapses the wedge sum $$X$$ to the $$i$$-th torus in the sum. Then we have a commutative diagram $$\require{AMScd} \begin{CD} H_2(M_g) @>{q_*}>> H^2(X) @>{(q_i)_*}>> H^2(T_i) \\ @V{\cong} VV @V{\cong}VV @V{\cong}VV \\ \mathbb{Z} @>>> \mathbb{Z}^g @>>> \mathbb{Z}. \end{CD}$$ Letting $$\sigma : \Delta^2 \to M_g$$ represent the generating class in $$H_2(M_g) \cong \mathbb{Z}$$, that $$\sigma$$ is not a boundary in $$H_2(M_g)$$ implies that $$q_i \circ q \circ \sigma$$ is itself not a boundary in $$H_2(T_i) \cong \mathbb{Z}$$. In particular, $$q_i \circ q \circ \sigma$$ generates $$H_2(T_i)$$ so we see that $$(q_i)_* \circ q_*$$ maps 1 to 1 along the bottom row of the diagram. Now, the isomorphism $$H_2(X) \cong \mathbb{Z}^g$$ comes from the isomorphism $$H_2(X) \cong \bigoplus_{i = 1}^g H_2(T_i)$$ which is induced by the inclusions $$T_i \hookrightarrow X$$ (Hatcher corollary 2.25). So if $$\sigma_i : \Delta^2 \to T_i$$ represents a generator of $$H_2(T_i)$$ then the map $$(q_i)_*$$ sends $$(0, \ldots, 0, \sigma_i, 0, \ldots, 0) \in H_2(X)$$ to $$\sigma_i$$. Along the bottom row of the diagram, if we regard $$\mathbb{Z}^g$$ as generated by the standard basis vectors $$\{e_1, \ldots, e_g\}$$, then $$(q_i)_*$$ acts as $$\varepsilon_i$$ for $$\varepsilon_i(e_j) := \delta_{ij}$$. We conclude that for each $$i$$ we have $$\varepsilon_i(q_*(1)) = 1$$ from which it follows that $$q_*(1) = (1, \ldots, 1) \in \mathbb{Z}^g \cong H_2(X)$$. Again apply the universal coefficient theorem for cohomology (noting that the Ext terms still vanish) to see that $$q^* : H^2(X) \to H^2(M_g)$$ maps each of the $$g$$ generators of $$H^2(X)$$ to the same generator of $$H^2(M_g)$$. In terms of the isomorphisms $$H^2(X) \cong \mathbb{Z}^g$$ and $$H^2(M_g) \cong \mathbb{Z}$$, we have concluded that $$q^*(e_i) = 1$$ for all $$i$$.

With a precise understanding of the action of $$q$$ on cohomology, we may return to the original commuting square. We adopt the notation of the OP here: let $$H^1(X)$$ have generators $$\{\alpha_i, \beta_i\}_{i = 1}^g$$ and $$H^2(X)$$ have generators $$\{\gamma_i\}_{i = 1}^g$$ with $$\alpha_i \beta_i = \gamma_i$$ and all other products of generators zero (note that we use here the known cup product structure on the torus and that we adopt the convention that multiplication denotes a cup product). Then that $$q^*$$ induces an isomorphism on first cohomology implies that $$H_1(M_g)$$ has generators $$\alpha_i' := q^*(\alpha_i)$$ and $$\beta_i' := q^*(\beta_i)$$. And the action of $$q^*$$ on $$H^2$$ means that $$H^2(M_g)$$ has generator $$\gamma' := q^*(\gamma_1) = \cdots = q^*(\gamma_g)$$. By our very first commuting square, we conclude that

Theorem: The cohomology ring $$\tilde{H}^*(M_g)$$ is generated by $$\{\alpha_i', \beta_i'\}_{i=1}^g \cup \{\gamma'\}$$ with $$\alpha_i'$$ and $$\beta_i'$$ of weight one, $$\gamma'$$ of weight two, and nonzero cup product relations $$\alpha_i' \beta_i' = \gamma'$$ for all $$i$$ (all other cup products of generators are zero).

The statement $H^*(\bigvee_i T^2) = \bigoplus_i \bigwedge[\alpha_i, \beta_i]$ is actually wrong. The formula you are thinking of only works for reduced cohomology in a given dimension and unfortunately reduced cohomology doesn't work with cup products since the zeroth reduced group is zero. We actually have that $H^*(\bigvee_i T^2)$ is the quotient of the direct sum, identifying the various $\mathbb{Z} \cong \bigwedge^0 [\alpha_i , \beta_i]$ together. (So its kinda like a wedge sum of graded abelian groups whatever that means). I think this is easy enough to see by considering the map from the disjoint union to the wedge sum.

To compute the ring structure on $M_g$ we first compute the abelian groups $H^i(M_g)$ and then use $q^*$ to determine the ring structure.

The cohomology groups of $M_g$ are actually the same as the homology groups by universal coefficients. Indeed the homology groups of $M_g$ are free abelian groups $H^0(M_g)=\mathbb{Z}$, $H^1(M_g)=\mathbb{Z}^{2g}$, $H^2(M_g)=\mathbb{Z}$ so the Ext terms vanish and we get isomorphisms $H^k(M_g)= \text{Hom}(H_k(M_g), \mathbb{Z}) = H_k(M_g)$. All other cohomology groups in higher dimensions are zero of course.

Because the cup product are maps $H^k(M_g) \times H^l(M_g) \to H^{k+l}(M_g)$ and the cohomology is zero above dimension two it follows that the only nontrivial cup product will be $H^1(M_g) \times H^1(M_g)$. (We also have the trivial cup product $H^0(M_g) \times H^l(M_g) \to H^l(M_g)$. This is the usual multiplication by integers since $H^0(M_g) = \mathbb{Z}$ consists of constant integer valued functions on the points of $M_g$.)

We can use $q$ to determine the cup product here, but I'll only give a sketch. Specifically we can show that $H^1(M_g) = \mathbb{Z}^{2g}$ has generators $q^*(\alpha_i)$, $q^*(\beta_i)$, $i=1, \dots, g$ and that $H^2(M_g) = \mathbb{Z}$ is generated by $q^*(\alpha_1 \cup \beta_1) = \cdots =\pm q^*(\alpha_g \cup \beta_g)$. Then because $q^*$ is a ring homomorphism it follows that $q^*(\alpha_i)\cup q^*(\beta_i) =q^*(\alpha_i \cup \beta_i)$, $q^*(\alpha_i) \cup q^*(\alpha_i) = q^*(\alpha_i \cup \alpha_i)=0$, and likewise $q^*(\beta_i)\cup q^*(\beta_i)= 0$.

In conclusion, $H^*(M_g)$ is the quotient of $H^*(\bigvee_i T^2)$, identifying the top $\mathbb{Z}$ summand and this has a natural ring structure (but no good notation as far as I'm aware).

• You managed to avoid getting into the details of everything that could have helped the OP.
– Pedro
Commented May 21, 2017 at 6:40
• I've edited for clarity. I'm sure there is still some argument to be had in showing $q^*$ maps generators to generators in the expected way, but this should put the OP on the right track. Commented May 22, 2017 at 3:19
• Yeah. I now have this same question, and it would be nice to see some more detail. Commented Apr 11, 2019 at 23:56
• Honestly, I don't know how you would make it more precise without a gluing diagram and even then you would need to use the cellular chain complex as a functor so $q$ gives an induced map between cellular chain complexes and I don't think thats a tool developed in hatcher at all (right?). Commented Apr 26, 2019 at 18:48