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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?

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    $\begingroup$ 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. $\endgroup$ – Justin Young Dec 12 '18 at 15:32
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    $\begingroup$ @JustinYoung Let $D^2$ be a two dimensional open disc. Is $(\mathbb{RP}^2 \times D^2) \setminus \{ \mathrm{point} \}$ homotopy equivalent to $\mathbb{RP}^2 \vee S^3$? $\endgroup$ – David E Speyer Jan 10 at 2:22
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    $\begingroup$ 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 $\mathbb{R}\mathbb{P}^2 \vee S^3$? $\endgroup$ – Hwang Jan 11 at 1:13
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    $\begingroup$ Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example. $\endgroup$ – Moishe Kohan Jan 12 at 3:03
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Lemma. Every finite CW complex is homotopy-equivalent to a smooth manifold .

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.

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  • $\begingroup$ It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do). $\endgroup$ – user98602 Jan 14 at 23:18
  • $\begingroup$ @MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now... $\endgroup$ – Moishe Kohan Jan 14 at 23:53

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