# Basic obstruction theory : where does the obstruction to uniqueness of lifting lie?

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 $$H^{n+1}(X,\pi_{n+1}(Y))$$" ?

• 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 $H^{m}(X,G)$. – Lee Mosher Sep 22 '19 at 15:25
• @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 – Maxime Ramzi Sep 22 '19 at 15:51
• And now I see that I mistyped. In my previous comment, change $H^m(X,G)$ to $H^{m+1}(X,G)$. – Lee Mosher Sep 22 '19 at 15:57
• 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) – Maxime Ramzi Sep 22 '19 at 17:05
• 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) – Maxime Ramzi Sep 22 '19 at 17:09

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.)

• 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. – Maxime Ramzi Sep 22 '19 at 21:28
• Although your answer is still interesting ! – Maxime Ramzi Sep 22 '19 at 21:28
• Oh, I see, I misread slightly. In any case the idea is the same although the example I gave isn't an example. – Qiaochu Yuan Sep 23 '19 at 5:04
• 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) – Maxime Ramzi Sep 23 '19 at 9:50
• (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 !) – Maxime Ramzi Sep 23 '19 at 9:58

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)

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.

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)