# Does the integral cohomology ring determine the ring structure with other coefficients?

Let $$A$$ be an abelian group. By the universal coefficient theorem, the cohomology group of a manifold with coefficient $$A$$ is determined by the integral cohomology group. How about the ring structure?

Let $$M_1$$ and $$M_2$$ be manifolds with $$H^*(M_1; \mathbb{Z}) \cong H^*(M_2; \mathbb{Z})$$ as a graded ring. Is it true that $$H^*(M_1; R) \cong H^*(M_2; R)$$ for any commutative ring $$R$$?

I guess the answer is no. What would be an example?

• The answer for CW complexes is no: $\mathbb RP^3$ and $\mathbb RP^2\vee S^3$ have isomorphic integral cohomology rings. I don't have manifold examples, though. Dec 12, 2018 at 15:32
• @JustinYoung Let $D^2$ be a two dimensional open disc. Is $(\mathbb{RP}^2 \times D^2) \setminus \{ \mathrm{point} \}$ homotopy equivalent to $\mathbb{RP}^2 \vee S^3$? Jan 10, 2019 at 2:22
• Could you explain a little more? The complement of a point in $\mathbb{R}\mathbb{P}^2 \times D^2$ is homotopy equivalent to $(\mathbb{R}\mathbb{P}^2 \times S^1) \cup S^1 \times D^2$. How do I see that this is homotopy equivalent to $\mathbb{R}\mathbb{P}^2 \vee S^3$? Jan 11, 2019 at 1:13
• Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example. Jan 12, 2019 at 3:03

Proof. (Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology" will contain all the necessary background information.) Every finite CW complex is homotopy-equivalent to the underlying space of a finite simplicial complex. Every finite simplicial complex is a subcomplex of the face-complex of a simplex. Thus, every finite simplicial complex embeds in $$R^n$$. (One can actually do a bit better: Each finite $$k$$-dimensional simplicial complex embeds in $$R^{2k+1}$$, this is a version of Whitney's embedding theorem in the PL category.) Now, let $$X\subset R^n$$ be a finite simplicial complex. Let $$N(X)$$ denote the regular neighborhood of $$X$$ in $$R^n$$. (The regular neighborhood of a subcomplex $$X$$ in a simplicial complex $$Y$$ is defined as follows. Take the 2nd barycentric subdivision $$Y''$$ of $$Y$$ and let $$N(X)$$ denote the subcomplex of $$Y''$$ consisting of simplices having nonempty intersection with $$X''$$.) Then $$N(R)$$ is homotopy-equivalent to $$|X|$$. (This is a general fact about regular neighborhoods of subcomplexes in simplicial complexes.) The regular neighborhood is a codimension zero PL manifold with boundary in $$R^n$$. Hence, the interior of $$N(X)$$ is an open subset in $$R^n$$ homotopy-equivalent to $$|X|$$. qed
Remark. One can do a bit better: If $$X$$ is a $$k$$-dimensional countable CW complex, then $$X$$ is homotopy-equivalent to a smooth $$2k$$-dimensional manifold. But you do not need this.
In view of this lemma, since you already have finite CW complexes answering your question ($$RP^3$$ and $$RP^2\vee S^3$$), you also have manifolds answering your question.