# Lifting cohomology-killing maps through the 3-sphere

In his first answer to this question, Jason deVito claimed that a map $f:X\to S^2$ kills $H^2$ if and only if it factors through the Hopf fibration $\pi:S^3\to S^2$. What's the justification for this claim?

(I can write the first couple of sentences in terms of CW complexes: to extend a lift $\tilde f$ defined on the $n-1$-skeleton to an $n$-cell $e_n$, note that $[f|_{\partial e_n}]=0\in \pi_{n-1}(S^2)$, lift a homotopy realizing that to $S^3$, so that we get $\tilde f$ homotopic to something mapping $\partial e_n$ into the fiber $S^1$. Then the altered $\tilde f$ will extend just if this defines a zero of $\pi_{n-1}(S^1)$. So we're only concerned when $n=2$. )

But I don't see how to get between this vanishing condition and that of $f^*$ killing $H^2$.

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I've got to run to class, and then I have a meeting. I should be able to post after that. – Jason DeVito Sep 3 '14 at 19:00
I don't know if it matters, but I now think this question is a duplicate of this one: math.stackexchange.com/questions/72166/…. Also, I've asked a follow up question about $\mathbb{H}P^n$ here:math.stackexchange.com/questions/920869/… – Jason DeVito Sep 5 '14 at 21:03

Let $\pi:S^{2n+1}\rightarrow \mathbb{C}P^n$ denote the projection map, giving $S^{2n+1}$ the structure of a principal $S^1$ bundle over $\mathbb{C}P^n$. Let $f:X\rightarrow \mathbb{C}P^n$ be any continuous map.

There is a function $\tilde{f}:X\rightarrow S^{2n+1}$ with $\pi\circ \tilde{f} = f$ iff $f^\ast:H^2(\mathbb{C}P^{n})\rightarrow H^2(S^{2n+1})$ is the $0$ map.

(I don't remember where, but I first learned of this statement for the $n=1$ case somewhere on MSE, but if I recall, there was no proof there. I don't claim the following proof is the most elegant, it's just what I came up with.)

First, the easy direction. If $\tilde{f}$ exists, then $f^\ast = \tilde{f}^\ast \circ \pi^{\ast}$. Since $\pi^{\ast}$ is trivial on $H^2$ for trivial reasons, $f^\ast$ must be trivial on $H^2$ as well.

Now, the fun direction. Assume $f^\ast$ is trivial on $H^2$. Consider the universal principal $S^1$ bundle $S^1\rightarrow S^\infty \rightarrow \mathbb{C}P^\infty$. The Hopf bundle $S^1\rightarrow S^{2n+1}\rightarrow \mathbb{C}P^n$ is classified by the inclusions map $i:\mathbb{C}P^n\rightarrow \mathbb{C}P^\infty.$

Now, consider the composition $i\circ f:X\rightarrow \mathbb{C}P^\infty$. Recall the following bijections. $$H^2(X;\mathbb{Z})\leftrightarrow [X,\mathbb{C}P^2]\leftrightarrow \{\text{Principal }S^1\text{ bundles over} X\}$$

The bijection between $H^2(X)$ and $[X,\mathbb{C}P^\infty]$ takes an element $g\in [X,\mathbb{C}P^\infty]$ to the element $g^\ast(z)\in H^2(X)$ where we once and for all choose a generator $z\in H^2(\mathbb{C}P^\infty)$.

Since, in our case, $i\circ f$ is trivial on $H^2$, it follows that the map $X\rightarrow \mathbb{C}P^\infty$ is homotopically trivial. In particular, if we pull back the universal bundle along $i\circ f$, we get the trivial $S^1$ bundle with total space $S^1\times X$. Call this projection map $\pi_2$.

But, this implies that $f$ pulls the Hopf bundle back to a trivial bundle. Being, a pull back bundle, there is a natural map $\hat{f}:f^\ast S^{2n+1}\rightarrow S^{2n+1}$ with the property that $f\circ p = \pi \circ \hat{f}$ where $p:f^\ast S^{2n+1}\rightarrow X$ is the projection. But $f^\ast S^{2n+1}$ is bundle isomorphic to $S^1\times X$, say by a bundle isomorphism $\phi$ which covers the identity.

(I can't seem to get commutative diagrams to work, but the idea is that you have four principal $S^1$ bundles in a row, starting with $\pi_2:S^1\times X \rightarrow X$, then $p:f^\ast S^{2n+1}\rightarrow X$, then $\pi:S^{2n+1}\rightarrow \mathbb{C}P^n$, then $\pi:S^\infty\rightarrow \mathbb{C}P^\infty.$)

Now, fix a point $s\in S^1$ and consider the inclusion $j:X\rightarrow S^1\times X$ with $j(x) = (s,x)$. (That is, pick a section).

Then $\tilde{f}$ is the composition $\hat{f}\circ \phi\circ j$. This follows because \begin{align*} \pi \circ \tilde{f} &= \pi \circ \hat{f}\circ \phi\circ j \\ &= f\circ p \circ \phi \circ j\\ &= f\circ Id_X \circ \pi_2\circ j\\ &=f\circ Id_X \circ Id_X\\ &= f \end{align*}

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This is just a fully detailed version of Omar's (perfectly fine) answer. I don't have any qualms with deleting it, if people would like me to. – Jason DeVito Sep 3 '14 at 20:40
Please, don't delete it! You wrote lots of details I skipped. – Omar Antolín-Camarena Sep 3 '14 at 20:41
I think in the displayed equation following "Recall the following bijections" you should have $\mathbb C P^{\infty}$ rather than $\mathbb CP^2$. – guy-in-seoul Sep 5 '14 at 21:23
@guy-in-seoul: I agree. If this question ever gets bumped again, someone should change it. (I don't think this minor of an edit is worth the bump.) – Jason DeVito Apr 18 '15 at 17:23

If you pullback the Hopf fibration along $f$, you get a principal $S^1$-bundle $P \to X$. If this bundle were trivial, it would have a section $X \to P$ and the composite $X \to P \to S^3$ would be the required lift of $f$ through the Hopf fibration.

So now we must show $f^\ast$ killing $H^2$ implies that $P \to X$ is the trivial bundle. Principal $S^1$-bundles on $X$ are classified by (homotopy classes of) maps $X \to BS^1$, the bundle $P \to X$ specifically, corresponds to the composite $X \xrightarrow{f} S^2 \xrightarrow{h} BS^1$ where $h$ is the map classifying the Hopf fibration; $P \to X$ is trivial if $h \circ f$ is null homotopic. Now, $BS^1$ is a $K(\mathbb{Z},2)$, so homotopy classes of maps $X \to BS^1$ are equivalently elements of $H^2(X, \mathbb{Z})$. And the Hopf fibration is classified by a map $h$ that corresponds to a generator of $H^2(S^2, \mathbb{Z}) \cong \mathbb{Z}$. So, $h \circ f$ is null homotopic if and only if $f^\ast$ kills $H^2(S^2,\mathbb{Z})$.

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Wow, we have essentially an identical proof. Way to beat me by 5 minutes! +1 – Jason DeVito Sep 3 '14 at 20:39
I'll leave it up, then. (Also, in your last line, I believe the $S^1$ should be an $S^2$.) – Jason DeVito Sep 3 '14 at 20:47
Thanks, @JasonDeVito, I've corrected it now. – Omar Antolín-Camarena Sep 3 '14 at 20:53