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Both $S^3\times \Bbb CP^\infty$ and $\left(S^1\times \Bbb CP^\infty\right)\big/\left(S^1\times \{x_0\}\right)$ have cohomology ring isomorphic to $\Bbb Z[a]\otimes \Lambda[b]$ with $|a|=2$ and $|b|=3$, as can be seen from Künneth and cellular cohomology. Thus, the cohomology ring structure can't distinguish the two spaces.

Hatcher says that the spaces can be distinguished using cohomology operations. I tried arguing somehow from the basic properties of Steenrod operations, but I couldn't figure it out.

I appreciate help.

iwriteonbananas
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    What about homotopy groups? – Pedro May 14 '16 at 17:31
  • Sure, @PedroTamaroff, but I would be more interested in seeing a solution using cohomology operations, as I am trying to study those right now. :) – iwriteonbananas May 15 '16 at 07:44
  • So, did you try computing stuff? – Pedro May 15 '16 at 08:06
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    Computing the first few homotopy groups of $S^3\times \Bbb CP^\infty$ isn't so hard. I think that the two spaces have different third homotopy group, but that's more of a guess as I'm not really sure how to compute homotopy groups for that quotient space.I can only show that it is simply connected and that $\Bbb Z$ surjects onto its second homotopy group. – iwriteonbananas May 15 '16 at 09:47

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First, consider $Y = \Bbb{CP}^\infty \times S^1$. Its cohomology ring over $\Bbb Z/2 =: \Bbb F$ is $\Bbb F[a] \otimes \Lambda[c]$ where $|a| = 2$ and $|c| = 1$. The Bockstein homomorphism can be seen to be trivial (as you say, cellular cohomology does the trick), and thus so is $\text{Sq}^1$. Now $\text{Sq}^2(ac) = \text{Sq}^2(a)c + \text{Sq}^1(a)\text{Sq}^1(c) + a\text{Sq}^2(c)$; because $|c| = 1$ the last term vanishes, and we have already discussed why the middle term vanishes; because $|a| = 2$, $\text{Sq}^2(a) = a^2$. Hence $\text{Sq}^2(ac) = a^2c$. If $X$ is your second space, the quotient map $p: Y \to X$ induces an injection $H^*(X) \to H^*(Y)$, sending $p^*(a) = a$ and $p^*(b) = ac$. Hence by naturality of Steenrod squares $\text{Sq}^2(b) = ab$.

Now consider $\Bbb{CP}^\infty \times S^3$. We want to calculate $\text{Sq}^2(b)$ here. The inclusion $S^2 \times S^3 \hookrightarrow \Bbb{CP}^\infty \times S^3$ induces an isomorphism on cohomology in degrees $\leq 5$, so by naturality we just need to understand what $\text{Sq}^2(b)$ is in the first space. A product $X \times Y$ of spaces with trivial Steenrod squares has trivial Steenrod squares. First note that because Steenrod squares are additive it suffices to see how the squares behave on elements in $H^i(X) \otimes H^j(Y) \subset H^{i+j}(X \times Y)$; and then by the Cartan formula it suffices inductively to see they vanish on $H^i(X)$ and $H^i(Y) \subset H^i(X \times Y)$. Now invoke the projection maps $\pi: X \times Y \to X$; by the vanishing of the Steenrod squares on $X$ and naturality, they vanish on these primitive classes in $X \times Y$.

Alternatively, we know that the Steenrod square commutes with the suspension isomorphism $H^*(S^2 \times S^3) \to H^{*+1}(\Sigma(S^2 \times S^3))$. As seen here, $\Sigma(X \times Y)$ is homotopy equivalent to $\Sigma X \vee \Sigma Y \vee \Sigma(X \wedge Y)$. In this case, this says that $\Sigma(S^2 \times S^3)$ is homotopy equivalent to $S^3 \vee S^4 \vee S^6$. This space has vanishing Steenrod squares, as does any wedge of spaces with vanishing Steenrod squares. (This is again by naturality, because we have projection maps $X \vee Y \to X$.)

Here's an alternate approach that Hatcher definitely did not intend, invoking characteristic classes of smooth manifolds; you can learn about these in Milnor and Stasheff's book on the subject.

Again we want to calculate $\text{Sq}^2(b)$ on $S^2 \times S^3$. Now, on a closed connected manifold, there are cohomology classes called the Wu classes, such that for $x \in H^{n-k}(M)$, $\text{Sq}^k(x) = x \smile v_k$. They are related to the Stiefel-Whitney classes of the manifold in that $\text{Sq}(v) = w$. Now, $S^2 \times S^3$ is parallelizable, so all of its Stiefel-Whitney classes vanish. This implies that $v=0$, and hence all of the Steenrod squares on this manifold (or any parallelizable manifold) vanish. Whence $\text{Sq}^2(b) = 0$, as desired.

In any case, we have now seen that $\text{Sq}^2(H^3(X)) \neq 0$, but $\text{Sq}^2(H^3(\Bbb{CP}^\infty \times S^3)) = 0$, proving the desired claim.

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  • The point, somehow, is that the Steenrod squares detect that $b \in H^2(X)$ "should split" as a product, even though you've killed the term it splits with, whereas it should be indecomposable in the other space. –  May 16 '16 at 08:07
  • This is the best answer ever - very instructive. Thank you. – iwriteonbananas May 16 '16 at 10:28
  • I am always utterly dumbfounded how people are so adept at these kinds of calculations. Sorry for being dense, but how is it obvious that the quotient map $p : Y \to X$ induces in injection on cohomology, and how do I determine it? Also, how do you know the Bockstein homomorphisms are trivial, and what does cellular cohomology have to do with it? –  Jun 15 '18 at 15:32
  • @Jeroen If I remember right, I spent the better part of a day working this out. It just always looks easier in the end. If you don't mind, give me a day to find some time to answer, and I'll think about your questions. –  Jun 15 '18 at 16:09
  • Of course, no problem. Thanks in advance. –  Jun 18 '18 at 09:19
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    @Jeroen So $Y$ is $S^1 \times \Bbb{CP}^\infty$ and $X$ is the space in the title. If one gives $S^1$ and $\Bbb{CP}^\infty$ the standard cell decompositions (whose cellular chain complexes have no differentials), the Kunneth theorem identifies the homology of the product with $H(C_(S^1) \otimes C_(\Bbb{CP}^\infty)$, but this tensor-product complex still has identically zero differentials (it's a tensor product of two complexes of such). To obtain the cellular chain complex of $X$, you just kill off a single generator of this chain complex. After dualizing, this surjection $C_* Y \to C_* X$... –  Jun 18 '18 at 14:07
  • ...becomes an injection. The cellular cochain complexes, though, are just the cohomology groups - again, no differential. To write down the map as a map of polynomial rings, just identify $H^(Y)$ as such a ring (by Kunneth), and then write down what generators of $H^(X)$ are in terms of the generators of $H^(Y)$, thinking of $H^(X)$ as a subgroup of $H^*(Y)$. –  Jun 18 '18 at 14:09
  • As for why the Bockstein is zero, I guess my point was is that one can determine entirely the behavior of the Bockstein by knowing that $0 \to H^(Y;\Bbb Z/2) \to H^(Y;\Bbb Z/4) \to H^(Y; \Bbb Z/2) \to 0$ is exact (so that there are no connecting homomorphisms, e.g., the Bockstein). However, for us, because the chain complex has no differential, we may identify each of $C^(Y;G)$ with $H^(Y;G)$, and the fact that $0 \to C^(Y;\Bbb Z/2) \to C^(Y;\Bbb Z/4) \to C^(Y;\Bbb Z/2) \to 0$ is exact implies that the same is true at the level of cohomology. –  Jun 18 '18 at 14:14
  • (I think all of this is constrained by comment space; if you come to the math chatroom and ping me I'm glad to talk more). –  Jun 18 '18 at 16:16
  • Thanks, this makes a lot of things clear. The fact that the cohomology of $X$ is generated by elements of degree $2$ and $3$ started to make sense when thinking geometrically about the 'killing' of a generator that you described. Looking back at the book I found that Hatcher does the same with quotients of the $n$-torus in Example 3.23. Though, I still find it hard to 'formalize' these thoughts, if you know what I mean. Perhaps that's just a case of a lack of solid intuition on my side. –  Jun 19 '18 at 09:35
  • Wait, this answer is wrong in at least one point: S^2 is not parallelizable, that's the hairy ball theorem. – dvitek Mar 15 '19 at 19:05