Let $X$ be a compact connected Hausdorff topological space.

We say $X$ is a cohomologicaly minimal space, briefly CM space, if $X$ satisfies the following property:

"For every proper subset $A\subset X$ with the inclusion map $i:A \to X$, $i^*$ is NOT a ring isomorphism between $H^*(X,\mathbb{Z})$ and $H^*(A,\mathbb{Z})$"

In the other word, $A$ is the only subset of $A$ whose inclusion gives a ring isomorphism cohomology.

Examples; All closed manifolds (see A closed manifold with a subset with the same ring cohomology )

Non Examples: Closed disc, figure 8,...etc.

My question

Is the product of two CM spaces, a CM space?


Let me say that $X$ has property $CM'$ if for every field $F$ and every point $a\in X$, the map $H_*(X\setminus a;F)\to H_*(X;F)$ is injective but not surjective. Then orientable closed manifolds have $CM'$, and $CM'$ implies $CM$. Now suppose that $X$ and $Y$ both have $CM'$. After noting that $$ (X\times Y)\setminus (a,b) = ((X\setminus a)\times Y) \cup (X\times (Y\setminus b)) $$ we can use the Kunneth isomorphism and the Mayer-Vietoris sequence to see that $X\times Y$ also has $CM'$. (I switched to using homology to ensure that Kunneth works in a straightforward way even if the spaces have infinitely generated homology.) I think that a similar approach can be made to work for non-orientable manifolds if we modify the definition of $CM'$ to distinguish between fields that do or do not have characteristic $2$. Thus, to make progress you should investigate whether there are any spaces that have $CM$ but not $CM'$.

Incidentally, contrary to your list of non-examples, I think that the figure 8 has $CM'$ (and therefore $CM$).

  • $\begingroup$ Thank you. Yes. It was my typos. Obviously figure 8 is a CM space. A question:Can the homology or cohomology version of Kunneth formula be applied for space which are not necessarily a CW complex? $\endgroup$ – Ali Taghavi Feb 1 '14 at 10:26
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    $\begingroup$ Let $X$ be two closed (non-zero-dimensional) manifolds connected by an interval. Then $X$ has $CM$ but not $CM'$ $\endgroup$ – Gabriel C. Drummond-Cole Feb 2 '14 at 13:32
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    $\begingroup$ One can use similar arguments to show the similar but incomparable statement that if $X\backslash \{a\}$ into $X$ and $Y\backslash \{b\}$ into $Y$ do not induce isomorphisms and $a$ and $b$ each has a contractible neighborhood, then the inclusion of $X\times Y\backslash (a,b)$ into $X\times Y$ cannot induce an isomorphism. $\endgroup$ – Gabriel C. Drummond-Cole Feb 7 '14 at 9:57

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