For singular cochain complexes we defined a cup-product: $$\cup: C^p(X,A;R)\otimes C^q(X,B;R)\to C^{p+q}(X,A\cup B;R)$$ $$(\alpha\cup\beta)(\sigma):=(-1)^{pq}\alpha(\sigma_{|[0,..,p]})\beta(\sigma_{|[p,..,p+q]}).$$ Here is $R$ a ring with unit, $X$ a topological space and $A,B\subseteq X$ and is $\sigma:\triangle ^n\to X$ a singular n-simplex. It is $\sigma_{|[0,..,p]}:\triangle ^p\to X,$ $$\sigma_{|[0,..,p]}(t_0,..,t_p)=\sigma(t_0,..,t_p,0..,0)$$ and $\sigma_{|[p,..,p+q]}:\triangle ^{p+q}\to X$ $$\sigma_{|[p,..,p+q]}(t_p,..,t_{p+q})=(0,..,t_p,..,t_{p+q}).$$ Then we defined the cup-product in singular cohomology $$\cup: H^p(X,A;R)\otimes H^q(X,B;R)\to H^{p+q}(X,A\cup B;R)$$ by $$\cup([\alpha],[\beta]):=[\alpha\cup\beta].$$ My questions are:

1)We already discussed singular homology. Is it possible to define a ring structure in a similar way on singular homology? Why we need cohomolgy at first? (At first the question was: Is the same definition possible for singular homology and if not, why not? This doesnt make sense)

This cup-product helps to define a ring structure on cohomology. I think we need this if we want to distinguish between topological spaces. 2)But why isn't homology and cohomology enough for that, why do we need the ring structure to distinguish between the topological spaces?

Best, ms.mop.

  • $\begingroup$ for your 2nd doubt $\mathbb{RP^3}$ and $\mathbb{RP^2}\vee S^3$ has same homology and cohomology , but they are not homotopically equivalent since there graded ring structures are different $\endgroup$ Nov 1, 2015 at 21:38
  • $\begingroup$ I dont understand your 1st doubt, I mean how do you want to define cup product or similar thing for chain complex?? $\endgroup$ Nov 1, 2015 at 21:42
  • $\begingroup$ thanks. you are right, I will try to describe more precisely what I really want to know. $\endgroup$
    – user197416
    Nov 1, 2015 at 21:45

1 Answer 1


1) Try and write down what you think it should be and see what goes wrong. What you actually get is a comultiplication $H_*(X) \to H_*(X) \otimes H_*(X)$, at least with field coefficients. I think things go wrong without field coefficients but I don't remember clearly why - it's been a while since I fiddled with this. E: There are much more intelligent thoughts about this on MathOverflow here.

2) Because of examples. $S^2 \times S^2$ is not homotopy equivalent to $\Bbb{CP}^2 \# \Bbb{CP}^2$ but they have the same homology. On the other hand, $$H^*(S^2 \times S^2) \cong \Bbb Z[x,y]/(x^2=y^2=0),$$ where $|x|=|y|=2$, and $$H^*(\Bbb{CP}^2\# \Bbb{CP}^2) \cong \Bbb Z[x,y]/(x^3=y^3=xy=0),$$ where again $|x|=|y|=2$. The first ring has degree 2 elements that square to zero while the second does not.

Indeed homology is nowhere near good enough to tell apart simply connected closed 4-manifolds, but the cohomology ring is a complete invariant.

  • $\begingroup$ That's a nice reference :) $\endgroup$ Nov 2, 2015 at 3:23
  • 3
    $\begingroup$ Remark: another example is $\Bbb{CP}^2$ and $S^2 \vee S^4$. It's not possible to distinguish them with just (co)homology groups. However, the cup product structure does indeed distinguish them - cup-square of $H^2$ in $\Bbb{CP}^2$ is a nontrivial generator of $H^4$ while is $0$ on $S^2 \vee S^4$. As a bonus, this in turn tells you that the Hopf map $S^3 \to S^2$ is homotopically nontrivial, hence $\pi_3(S^2)$ is nonzero. One can in general use the cup-squared structure on maps $S^{4n-1} \to S^{2n}$ to investigate $\pi_{4n-1}(S^{2n})$. This is known as the Hopf invariant. $\endgroup$ Nov 2, 2015 at 10:43
  • $\begingroup$ thank you!!! very helpful!! $\endgroup$
    – user197416
    Nov 7, 2015 at 11:32
  • $\begingroup$ For any ring you get a comultipliation on chains, but when you pass to homology, there can be problems with passing the homology of the tensor product to the tensor product of the homologies. $\endgroup$ Feb 19, 2016 at 17:38

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