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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.