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edited Feb 17 2011 at 10:24 |
and $f:X\to X$ is given by $f(x,y):= (x,y + u(x))$, using that EM-spaces are abelian groups). There are also self-maps of finite simply connected complexes that are the identity on homology, but not on homotopy, see Diarmuid Crowleys answer to http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure/55609#55609Construct two simply-connected CW complexes $X$ and $Y$ such that $H^* (X;F) \cong H^* (Y;F)$ for any field, as rings and modules over the Steenrod algebras, but which are not homotopy equivalent. EDIT: Appropriate Moore spaces do the job, see Eric Wofseys answer. Let $f: X \to Y$ be a map of CW-complexes. Assume that $[T,X] \to [T;Y]$ is a bijection for each finite CW complex $T$ ($[T,X]$ denotes free homotopy classes). What assumptions are sufficient to conclude that $f$ is a weak homotopy equivalence? EDIT: the answer has been given by Tyler Lawson, see below. Do there exist spaces $X,Y,Z$ and a homotopy equivalence $X \times Y \to X \times Z$, without $Y$ and $Z$ being homotopy equivalent? Can I require these spaces to be finite CWs? EDIT: without the finiteness assumptions, this question was ridiculously simple. I am looking forward for your answers. EDIT: so far, I have gotten great answers, but mostly for the specific questions I asked. My intention was to create a larger list of counterexamples. So, feel free to mention your favorite strange spaces and maps.
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edited Feb 14 2011 at 11:15 |
Do you know examples of fibrations $F \to E \to X$, such that the integral homology of all three spaces is finitely generated (so that the Euler numbers are defined) and such that the Euler number is not multiplicative, i.e. $\chi(E) \neq \chi(F) \chi(X)$? Remark: is $X$ is assumed to be simply-connected, then the Euler number is multiplicative (absolutely standard). Likewise, if $X$ is a finite CW complex and $F$ is of finite homological type (less standard, but a not so hard exercise). So any counterexample would have to be of infinite type. The above fibration $BSL_2 (Z) \to BZ/12$ is a counterexample away from the primes $2,3$, but do you know one that does the job in all characteristics?. Of course, the ordinary Euler number is the wrong concept here.
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Post Made Community Wiki by S. Carnahan♦
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occurred Feb 14 2011 at 5:40
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edited Feb 14 2011 at 3:48 |
In this thread
http://mathoverflow.net/questions/53036/books-you-would-like-to-read-if-somebody-would-just-write-them/53059#53059,
I expressed my desire for a book with the title "(Counter)examples in Algebraic Topology".
My reason for doing so was that while the abstract formalism of algebraic topology is very well-explained in many textbooks and while most graduate students are fond of the general machinery, the study of examples is somehow neglected. I am looking for examples that explain why certain hypotheses are necessary for theorems to hold. The books by Hatcher and Bredon contain some interesting stuff in this direction, and there is Neil Strickland's bestiary, which is mainly focussed on positive knowledge.
To convey an idea of what I am after, here are a few examples from my private ''counterexamples in algebraic topology'' list. Some are surprising, some less so.
The abelianization of $SL_2 (Z)$ is $Z/12$, the map $BSL_2(Z) \to BZ/12$ is a homology equivalence to a simple space. But it is not a Quillen plus construction, since the the homotopy fibre is $BF_2$ (free group on $2$ generators), hence not acyclic. See http://mathoverflow.net/questions/43726/the-free-group-f-2-has-index-12-in-sl2-mathbbz/43741#43741. Maps $f:X \to Y$ which are homology equivalences, the homotopy groups are abstractly isomorphic, but though, $f$ is not a homotopy equivalence (a number of examples has been given in the answers to these questions:
http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409,
http://mathoverflow.net/questions/4665/are-there-pairs-of-highly-connected-finite-cw-complexes-with-the-same-homotopy-gr). Self-maps of simply-connected spaces $X$ which are the identity on homotopy, but not on homology (let $X=K(Z;2) \times K(Z;4)$, $u:K(Z;2) \to K(Z;4)$ be the cup square,
and $f:X\to X$ is given by $f(x,y):= (x,y + u(x))$, using that EM-spaces are abelian groups). The stabilization map $B \Sigma_{\infty} \to B \Sigma_{\infty}$ induces a bijection on free homotopy classes $[X, B \Sigma_{\infty}]$ for each finite CW space $X$. However, it is not a homotopy equivalence (not a $\pi_1$-isomorphism). The fibration $S^1 \to B \mathbb{Q} \to B \mathbb{Q}/\mathbb{Z}$ is classified by a map $f:B \mathbb{Q}/\mathbb{Z} \to CP^{\infty}$, which can be assumed to be a fibration with fibre $B \mathbb{Q}$. Now let $X_n$ be the preimage of the n-skeleton of $CP^{\infty}$. Using the Leray-Serre spectral sequence, we can compute the integral
homology of $X_n$ and, by the universal coefficient theorem, the homology of field coefficients. It turns out that this is finitely generated for any field,
and so we can define the Euler characteristic in dependence of the field. It is not independent of the field in this case (the reason is of course that the integral homology of $X_n$ is not finitely generated). The compact Lie groups $U(n)$ and $S^1 \times SU(n)$ are diffeomorphic, their classifying spaces have isomorphic cohomology rings and homotopy groups, but they the classifying spaces are not homotopy equivalent (look at Steenrod operations).
Question: Which examples of spaces and maps of a similar flavour do you know and want to share with the other MO users?
To focus this question, I suggest to stay in the realm of algebraic topology proper. In other words:
The properties in question should be homotopy invariant properties of spaces/maps. This includes of course fibre bundles, viewed as maps to certain classifying spaces. Let us talk about spaces of the homotopy type of CW complexes, to avoid that a certain property fails for point-set topological reasons. This excludes the kind of examples from the famous book "Counterexamples in Topology". The examples should not be "counterexamples in group theory" in disguise. Any ugly example of a discrete group $G$ gives an equally ugly example of a space $BG$.
Same applies to rings via Eilenberg Mac-Lane spectra. I prefer examples from unstable homotopy theory.
To get started, here are some questions whose answer I do not know:
Construct two simply-connected CW complexes $X$ and $Y$ such that $H^* (X;F) \cong H^* (Y;F)$ for any field, as rings and modules over the Steenrod algebras, but which are not homotopy equivalent. Let $f: X \to Y$ be a map of CW-complexes. Assume that $[T,X] \to [T;Y]$ is a bijection for each finite CW complex $T$ ($[T,X]$ denotes free homotopy classes). What assumptions are sufficient to conclude that $f$ is a weak homotopy equivalence? Do there exist spaces $X,Y,Z$ and a homotopy equivalence $X \times Y \to X \times Z$, without $Y$ and $Z$ being homotopy equivalent? Can I require these spaces to be finite CWs?
I am looking forward for your answers.
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asked Feb 14 2011 at 0:14 |
Counterexamples in algebraic topology?
In this thread
http://mathoverflow.net/questions/53036/books-you-would-like-to-read-if-somebody-would-just-write-them/53059#53059,
I expressed my desire for a book with the title "(Counter)examples in Algebraic Topology".
My reason for doing so was that while the abstract formalism of algebraic topology is very well-explained in many textbooks and while most graduate students are fond of the general machinery, the study of examples is somehow neglected. I am looking for examples that explain why certain hypotheses are necessary for theorems to hold. The books by Hatcher and Bredon contain some interesting stuff in this direction, and there is Neil Strickland's bestiary, which is mainly focussed on positive knowledge.
To convey an idea of what I am after, here are a few examples from my private ''counterexamples in algebraic topology'' list. Some are surprising, some less so.
The abelianization of $SL_2 (Z)$ is $Z/12$, the map $BSL_2(Z) \to BZ/12$ is a homology equivalence to a simple space. But it is not a Quillen plus construction, since the the homotopy fibre is $BF_2$ (free group on $2$ generators), hence not acyclic. See http://mathoverflow.net/questions/43726/the-free-group-f-2-has-index-12-in-sl2-mathbbz/43741#43741. Maps $f:X \to Y$ which are homology equivalences, the homotopy groups are abstractly isomorphic, but though, $f$ is not a homotopy equivalence (a number of examples has been given in the answers to these questions:
http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409,
http://mathoverflow.net/questions/4665/are-there-pairs-of-highly-connected-finite-cw-complexes-with-the-same-homotopy-gr). Self-maps of simply-connected spaces $X$ which are the identity on homotopy, but not on homology (let $X=K(Z;2) \times K(Z;4)$, $u:K(Z;2) \to K(Z;4)$ be the cup square,
and $f:X\to X$ is given by $f(x,y):= (x,y + u(x))$, using that EM-spaces are abelian groups). The stabilization map $B \Sigma_{\infty} \to B \Sigma_{\infty}$ induces a bijection on free homotopy classes $[X, B \Sigma_{\infty}]$ for each finite CW space $X$. However, it is not a homotopy equivalence (not a $\pi_1$-isomorphism). The fibration $S^1 \to B \mathbb{Q} \to B \mathbb{Q}/\mathbb{Z}$ is classified by a map $f:B \mathbb{Q}/\mathbb{Z} \to CP^{\infty}$, which can be assumed to be a fibration with fibre $B \mathbb{Q}$. Now let $X_n$ be the preimage of the n-skeleton of $CP^{\infty}$. Using the Leray-Serre spectral sequence, we can compute the integral
homology of $X_n$ and, by the universal coefficient theorem, the homology of field coefficients. It turns out that this is finitely generated for any field,
and so we can define the Euler characteristic in dependence of the field. It is not independent of the field in this case (the reason is of course that the integral homology of $X_n$ is not finitely generated). The compact Lie groups $U(n)$ and $S^1 \times SU(n)$ are diffeomorphic, their classifying spaces have isomorphic cohomology rings and homotopy groups, but they are not homotopy equivalent (look at Steenrod operations).
Question: Which examples of spaces and maps of a similar flavour do you know and want to share with the other MO users?
To focus this question, I suggest to stay in the realm of algebraic topology proper. In other words:
The properties in question should be homotopy invariant properties of spaces/maps. This includes of course fibre bundles, viewed as maps to certain classifying spaces. Let us talk about spaces of the homotopy type of CW complexes, to avoid that a certain property fails for point-set topological reasons. This excludes the kind of examples from the famous book "Counterexamples in Topology". The examples should not be "counterexamples in group theory" in disguise. Any ugly example of a discrete group $G$ gives an equally ugly example of a space $BG$.
Same applies to rings via Eilenberg Mac-Lane spectra. I prefer examples from unstable homotopy theory.
To get started, here are some questions whose answer I do not know:
Construct two simply-connected CW complexes $X$ and $Y$ such that $H^* (X;F) \cong H^* (Y;F)$ for any field, as rings and modules over the Steenrod algebras, but which are not homotopy equivalent. Let $f: X \to Y$ be a map of CW-complexes. Assume that $[T,X] \to [T;Y]$ is a bijection for each finite CW complex $T$ ($[T,X]$ denotes free homotopy classes). What assumptions are sufficient to conclude that $f$ is a weak homotopy equivalence? Do there exist spaces $X,Y,Z$ and a homotopy equivalence $X \times Y \to X \times Z$, without $Y$ and $Z$ being homotopy equivalent? Can I require these spaces to be finite CWs?
I am looking forward for your answers.
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