Suppose $X$ and $Y$ are finite cell complexes (then I may assume each point has a contractible neighborhood). Suppose we want to compute the cohomology ring structure of $X\vee Y$ (i.e. the wedge sum of both spaces). My lecturer proceeded as follows: use the cohomology Mayer Vietoris sequence. An easy argument shows that $$H^* (X\vee Y)\cong H^*(X)\times H^*(Y)$$ for $^*>0$; and $$H^0 (X\vee Y)\cong H^0(X)\times H^0(Y) / \langle(1_X,1_Y)\rangle$$ where $1_X$ denotes the identity of $H^*(X)$ and similarly for $1_Y$. Then he concluded that $$H^* (X\vee Y)\cong H^*(X)\times H^*(Y) / \langle(1_X,1_Y)\rangle$$ and said that "the cohomology ring of the wedge sum is the product of the cohomology rings of the factors modulo some relation in grading zero". But I don't understand such isomorphism. In fact, $ \langle(1_X,1_Y)\rangle$ is an abelian group (it is not a (graded) ideal of the (graded) product ring). So what sense does it make the quotient as a ring?

I am aware that one could use reduced cohomology , but I want to work this out by using the absolute version....

Thank you


1 Answer 1


The right way to think of $H^*(X\vee Y)$ is not as a quotient of $H^*(X)\times H^*(Y)$ (which, as you point out, does not have a natural ring structure) but rather as a subring. Indeed, this is what the Mayer-Vietoris sequence actually gives you, since the natural map is $H^*(X\vee Y)\to H^*(X)\times H^*(Y)$, not the other way around.

Specifically, $H^*(X\vee Y)$ is the subring of $H^*(X)\times H^*(Y)$ which contains everything in positive degrees and in degree $0$ consists of the set of pairs $(f,g)\in H^0(X)\times H^0(Y)$ such that $f(x_0)=g(y_0)$, where $x_0$ and $y_0$ are the basepoints of $X$ and $Y$ (considered as $0$-cycles on $X$ and $Y$). Concretely, if $X$ has $m$ components and $Y$ has $n$ components so $H^0(X)\cong\mathbb{Z}^m$ and $H^0(Y)\cong\mathbb{Z}^n$, with the first coordinate in each corresponding to the component of the basepoint, then $H^0(X\vee Y)$ is the subring of $\mathbb{Z}^{m+n}$ consisting of all tuples whose first and $(m+1)$st coordinates are equal (which is isomorphic to $\mathbb{Z}^{m+n-1}$, by just removing one of those coordinates as redundant).

More conceptually, $H^*(X\vee Y)$ is the subring of $H^*(X)\times H^*(Y)$ consisting of all pairs $(a,b)$ such that $i^*(a)=j^*(b)$, where $i:\{*\}\to X$ and $j:\{*\}\to Y$ are the inclusions of the basepoints. Indeed, this is exactly the description given by the Mayer-Vietoris sequence. Since $H^*(\{*\})$ is trivial in positive degrees, this means $i^*(a)=j^*(b)$ is always true in positive degrees, and the only restriction is in degree $0$ as described above.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .