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This is a question about a remark someone said to me without giving much precision.

Suppose you have two nice spaces $X,Y$ and are trying to build a map $X\to Y$ with certain nice properties. Suppose for simplicity (no pun intended) that $Y$ is a simple space, that is $\pi_1(Y)$ acts trivially on $\pi_n(Y)$ for all $n$.

Then one way to do this is to decompose $Y$ into a Postnikov tower $\dots \to Y_2\to Y_1$. As $Y$ is simple, we can choose each $Y_{n+1}\to Y_n$ to be a principal fibration.

I was told that the "obstruction to the existence of a lift $f:X\to Y_n$ to $\tilde f : X\to Y_{n+1}$ is a cohomology class in $H^{n+2}(X,\pi_{n+1}(Y))$" and that the "obstruction to the uniqueness of such a lift lies in $H^{n+1}(X,\pi_{n+1}(Y))$".

I understand the first bit : indeed if we look at a delooping of the fibration $Y_{n+1}\to Y_n$ we see that it is of the form $X_{n+1}\to X_n \to K(\pi_{n+1}(X), n+2)$, therefore if we hace $f: X\to Y_n$, it lifts (up to homotopy) to $Y_{n+1}$ if and only if it is sent to $0$ in $[X, K(\pi_{n+1}(X), n+2)] = H^{n+2}(X, \pi_{n+1}(Y))$, so the obstruction is the class of the pushforward of $f$ in $H^{n+2}(X, \pi_{n+1}(Y))$.

I have more trouble with the second bit, though. I understand that it is related to the fact that the fiber of the fibration $Y_{n+1}\to Y_n$ is $K(\pi_{n+1}(Y), n+1)$, so if somehow I could "subtract" maps I would definitely get the obstruction where I was told it was; but without that I seem to be stuck :

I have two maps $f_1,f_2 : X\to Y_{n+1}$ that lift $f:X\to Y_n$, what do I do with them ? How do I extract a map $X\to $fiber ? Or perhaps $X\to \mathrm{hofib}$ ?

I thought of focusing on one of the two maps, say $f_1$, fixing $f$ and seeing that a homotopy $p\circ f_1\to f$ (where $p: Y_{n+1}\to Y_n$) gives me a map $X\to$ some space that looks like the homotopy fiber, but I can't make that precise (I would want something like "the homotopy fiber over a point that moves along with $X$")

So my question is :

What is meant by "the obstruction to uniqueness of lifing lies in $H^{n+1}(X,\pi_{n+1}(Y))$" ?

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  • $\begingroup$ If I were to pose this simpler problem, would you know what to do? And/or would knowing what to do be sufficient to you to settle your actual problem? Namely, given $K = K(G,m+1)$ and given two maps $g_0,g_1 : X \to K$, show that the obstruction to a homotopy from $g_0$ to $g_1$ lives in $H^{m}(X,G)$. $\endgroup$ – Lee Mosher Sep 22 '19 at 15:25
  • $\begingroup$ @LeeMosher : Right now I don't see, but I'll try to think about it. As far as I can see the simpler problem is the question of a lift $X\to K^2$ to $X\to K^I$. Perhaps there's a way to see that $K^I\to K^2$ is a principal fibration ? I'll see what I can find $\endgroup$ – Maxime Ramzi Sep 22 '19 at 15:51
  • $\begingroup$ And now I see that I mistyped. In my previous comment, change $H^m(X,G)$ to $H^{m+1}(X,G)$. $\endgroup$ – Lee Mosher Sep 22 '19 at 15:57
  • $\begingroup$ Ok here's a better solution to the simpler problem : $K$ is a loop space, so we have a natural multiplication map $K^2\to K$. Let's look more generally at the homotopy fiber of $\Omega Y^2\to \Omega Y$ : a point in it consists in a couple $(\gamma_1, \gamma_2)$ of loops together with a path $\gamma_1\gamma_2 \to *$ ($*$ being the constant loop, where $Y$ is based). But such a path can equally be interpreted as a path homotopy $\gamma_1\to \gamma_2^{-1}$. Since all these things happen at the level of parametrizations of $I$, this should be continuous, and in fact a homotopy equivalence. (1/2) $\endgroup$ – Maxime Ramzi Sep 22 '19 at 17:05
  • $\begingroup$ But inversion is a homotopy equivalence (in fact a homeomorphism $\Omega Y \to \Omega Y$ so in fact the homotopy fiber is the space of couples $(\gamma_1, \gamma_2)$ together with a homotopy $\gamma_1\to \gamma_2$, that is, the homotopy fiber is precisely $(\Omega Y)^I$, with the correct "inclusion" map (if I change $(x,y)\mapsto xy$ to $(x,y)\mapsto xy^{-1}$ ). This sounds more reasonable than what I had before, and in fact it corresponds more closely to the idea of subtracting maps in cohomology. It follows that we have a fiber sequence $K^I \to K^2 \to K$ (2/ more than 2 actually) $\endgroup$ – Maxime Ramzi Sep 22 '19 at 17:09
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I'll write $B^n A$ for $K(A, n)$. Given that there exists a lift, the space of lifts is the space of homotopy sections of the homotopy pullback of the bundle $Y_{n+1} \to Y_n$ to $X$ (this follows just from the universal property of the homotopy pullback). The bundle $Y_{n+1} \to Y_n$ is a principal $B^{n+1} \pi_{n+1}(Y)$-bundle whose pullback to $X$ admits a section, hence which is trivializable over $X$. The space of sections of the trivial bundle is the space of functions $[X, B^{n+1} \pi_{n+1}(Y)]$, whose $\pi_0$ is $H^{n+1}(X, \pi_{n+1}(Y))$, and so the space of sections of any trivializable bundle is naturally a torsor over this space.

(This is a special case of a very general pattern that is straightforward when stated abstractly but surprisingly hard to spot: if $a, b$ are isomorphic, the space of isomorphisms between them is naturally a torsor over the automorphism group of either. "Trivializable" means isomorphic to the trivial bundle, and the space of sections of a principal bundle can be naturally identified with the space of isomorphisms to the trivial bundle.)

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  • $\begingroup$ No I really did mean Postnikov Tower. Perhaps there is also an obstruction theory using Whitehead towers, but I meant Postnikov : $Y_1$ is $K(\pi_1, 1)$, $Y_2$ has two (possibly) nontrivial homotopy groups etc. $\endgroup$ – Maxime Ramzi Sep 22 '19 at 21:28
  • $\begingroup$ Although your answer is still interesting ! $\endgroup$ – Maxime Ramzi Sep 22 '19 at 21:28
  • $\begingroup$ Oh, I see, I misread slightly. In any case the idea is the same although the example I gave isn't an example. $\endgroup$ – Qiaochu Yuan Sep 23 '19 at 5:04
  • $\begingroup$ I still have some things to unwind from your answer : for instance what is a principal something-bundle when something is not a group, why is $Y_{n+1}\to Y_n$ such a bundle, and why are these bundles stable under homotopy pullback; why are such bundles with a section trivializable (for principal $G$-bundles that's quite clear, but well I don't know what meaning you give to bundle here). Once I figure all this out I'll get it (right now I only understand the idea of what you're saying but the details are still elusive) $\endgroup$ – Maxime Ramzi Sep 23 '19 at 9:50
  • $\begingroup$ (if you can explain what bundles are in this context, or if you can give some reference where I can read about it, I'd be very grateful !) $\endgroup$ – Maxime Ramzi Sep 23 '19 at 9:58
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Let me write another answer to try to understand this from another point of view and using things I learned recently.

Let me assume that $X$ is simply-connected and very nice.

We're back at our stage where we have $X\to Y_n$ and we're asking when we can lift it (uniquely) to $X\to Y_{n+1}$.

By taking the homotopy pullback of $Y_{n+1}\to Y_n$ along $X\to Y_n$ we get a fibration $P\to X$ with fiber $K(\pi_{n+1}(Y),n+1)$. Let me write $\pi = \pi_{n+1}(Y)$ for simplicity of notation.

The existence of a lift is equivalent to the existence of a section of that fibration, and then the space of lifts is equivalently the space of sections.

But now this fibration $P\to X$ is also a (Kan) fibration of $\infty$-groupoids, so in particular I can see it as an $\infty$-functor $X\to \mathcal S$ ($\mathcal S$ is the $\infty$-category of spaces).

Up to equivalence I may assume that this functor takes a constant value equal to $K(\pi,n+1)$ on objects.

It follows that we get a functor $\chi : X\to BAut(K(\pi,n+1))$ where by this I denote the full sub-$\infty$-groupoid of $\mathcal S^{\simeq}$ on $K(\pi,n+1)$, which classifies our fibration $P\to X$. Note that the space of sections of that fibration is the space of maps from the constant functor equal to a point to that functor

So now I have to examine what this space $BAut(K(\pi,n+1))$ is.

It has one vertex, with mapping space $Map^\simeq (K(\pi,n+1),K(\pi,n+1))$

Therefore its $\pi_0$ is trivial and $\pi_k$ of it is $\pi_0\Omega^k$ of it. $\Omega$ of it is $Map^\simeq(K(\pi,n+1),K(\pi,n+1))$ so we're left with computing the homotopy groups of that thing.

Its $\pi_0$ is a subset of $H^{n+1}(K(\pi,n+1),\pi) = \hom(\pi,\pi)$ and it's easy to check that it's precisely $Aut(\pi)$.

Then its $\pi_k$ at $id_\pi$ is $\pi_0\Omega^kMap(K(\pi,n+1),K(\pi,n+1)) = \pi_0Map(K(\pi,n+1), K(\pi,n+1-k)) = H^{n+1-k}(K(\pi,n+1),\pi) =$ $\pi$ if $k=n+1$, $0$ else.

So $BAut(K(\pi,n+1))$ has two homotopy groups : a $\pi_1= Aut(\pi)$, and a $\pi_{n+2} = \pi$.

Now since $X$ is simply-connected, the map $X\to BAut(K(\pi,n+1))$ lifts to $X\to K(\pi,n+2)$ because of the fiber sequence $K(\pi,n+2)\to BAut(K(\pi,n+1)) \to K(Aut(\pi),1)$

So our fibration is classified by some cohomology class $\alpha \in H^{n+2}(X,\pi)$. More precisely, the space of fibrations of this type is $Map(X,BAut(K(\pi,n+1)))\simeq Map(X,K(\pi,n+2))$

If the fibration has a section, then we get a map of fibrations from $X\to X$ to $P\to X$ which translates to a map of functors $cst_* \to \chi$ and it follows that $\alpha = 0$.

Therefore the obstruction to lifting is indeed a cohomology class $\alpha \in H^{n+2}(X,\pi)$ (not sure how useful that construction is, though)

Now how about uniqueness ?

Well if our fibration is known to be trivial, that is, of the form $K(\pi,n+1)\times X$, then the space of sections is $Map(X,K(\pi,n+1))$ (whose $\pi_0$ is indeed $H^{n+1}(X,\pi)$)

More generally, the existence of a map $cst_*\to \chi$ should guarantee the existence of $cst_{K(\pi,n+1)}\to \chi$ because $K(\pi,n+1)$ is a "group". But one would need $\chi$ to be a functor with values in $K(\pi,n+1)$ modules, or, equivalently, $P\to X$ to have a compatible $K(\pi,n+1)$ action. Is that clear ?

It seems to me like this amounts to claiming that any (split ?) fibration $G\to P\to X$ with $G$ an $E_1$-space automatically inherits a (canonical) $G$-action on $P$ which induces the right action on $G$.

I'll have to think about that, it might be right, but it doesn't seem to be that obvious.

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Here's a more lowbrow point of view, with, I think, no gap - as opposed to my previous attempt at an answer .

The idea is that instead of separating uniqueness and existence, we consider the two similarly, by simply using a relative version of the obstruction to lifting.

So suppose we have a map $X\to Y_n$, a map (think of it as a cofibration) $A\to X$ together with a lift of $A\to X\to Y_n$ to $Y_{n+1}$. That is, we have a commutative square

$\require{AMScd}\begin{CD} A @>>> Y_{n+1} \\ @VVV @VVV \\ X @>>> Y_n\end{CD}$

What is the obstruction to filling this diagram with a diagonal map $X\to Y_{n+1}$ ? (up to homotopy, say)

So this is a relative version of the obstruction to existence, but we'll see that it can be adapted to provide obstruction to unicity as well.

Just as before, if $Y$ is good enough, the Postnikov tower consists of principal fibrations, and in fact, $Y_{n+1}\to Y_n$ is the pullback of the fibration $*\to K(\pi_{n+1}(Y), n+2)$ along some map $Y_n\to K(\pi_{n+1}(Y),n+2)$

It follows that we actually have a larger diagram

$\require{AMScd}\begin{CD} A @>>> Y_{n+1} @>>> *\\ @VVV @VVV @VVV\\ X @>>> Y_n @>>> K(\pi_{n+1}(Y),n+2)\end{CD}$

Now finding a diagonal filler in the original diagram is the same as finding a diagonal filler in

$\require{AMScd}\begin{CD} A @>>> *\\ @VVV @VVV\\ X @>>> K(\pi_{n+1}(Y),n+2)\end{CD}$

by pullback properties.

But now that diagram corresponds to a map $cofiber(A\to X)\to K(\pi_{n+1}(Y),n+2)$, and if that map is not nullhomotopic, there is no lifting. In fact, the obstruction to getting a diagonal filler is precisely this map; that is, it's a cohomology class in $H^{n+2}(Cofiber(A\to X), \pi_{n+1}(Y))$.

Now how does this help us ? Well in the $A=\emptyset$ scenario, we get back the non-relative lifting problem, that is, the obstruction to lifting is a cohomology class in $H^{n+2}(X, \pi_{n+1}(Y))$, as initially announced.

What the relative version buys us, though, is that if we take $A= X\times \partial I \to X\times I$ as our cofibration, the problem of finding a diagonal filler is essentially the problem of uniqueness of a lift. That is, if we have two lifts of the same map, they correspond to a diagram

$\require{AMScd}\begin{CD} X\times \partial I @>>> Y_{n+1} \\ @VVV @VVV \\ X\times I @>>> Y_n\end{CD}$

and a diagonal filler is precisely a homotopy between the two. So the obstruction to such a filler is precisely the obstruction to uniqueness, so it's a cohomology class in $H^{n+2}(X\times I/ X\times \partial I, \pi_{n+1}(Y))= H^{n+1}(X,\pi_{n+1}(Y))$, which is what we wanted.

(this last equality can be seen from the answer here. The idea is simply that $(X\times I)/(X\times \partial I) = \Sigma(X_+)$, so the reduced cohomology is that of $X$, shifted by one degree)

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