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I'm trying to learn about coexact sequences for maps. On p. 445 of Bredon's "Topology and Geometry", Bredon defines a sequence

$A \xrightarrow{f} B \xrightarrow{g} C$

of pointed topological spaces to be coexact if, for each pointed space $Y$, the sequence of sets (pointed homotopy classes)

$[C;Y] \xrightarrow{g^\sharp}[B;Y]\xrightarrow{f^\sharp}[A;Y]$

is exact, i.e., $\text{im}(g^\sharp)=(f^\sharp)^{-1}(*)$.

Question: Suppose that the following sequence of pointed spaces

$* \to A \xrightarrow{f} B \to *$

is coexact, where $*$ denotes the one-point space. (Here coexactness means that each of the two short subsequences of two maps is coexact.) Does this imply that $f$ is a homotopy equivalence? I would really appreciate a proof/counterexample and/or reference.

Motivation: At the bottom of p. 62 of Conley's "Isolated Invariant Sets and the Morse Index", Conley seems to assert this. (In case it matters, I believe his coexact sequence is a special case of Barratt-Puppe on p.447 of Bredon.) Conley cites Spanier's "Cohomology Theory for General Spaces" as a source for this assertion, but after looking at that paper I'm wondering if Conley made a mistake and actually meant to cite Spanier's book "Algebraic Topology." I looked at Ch. 7.1 of Spanier's book and found a treatment very similar to Bredon's, but I didn't find the statement I was looking for. I have also tried proving this from the definition, but I haven't figured it out. I'm also feeling paranoid that the statement might not be true since I haven't managed to find a source/proof for this claim.

Update 1: Using LordSharktheUnknown's suggestion, take $Y = A$ and consider the identity map $\text{id}_A$. This gets pulled back under $*\to A$ to a constant map. By coexactness, therefore there exists $g: B \to A$ with $f^\sharp g = g \circ f$ homotopic to $\text{id}_A$. Hence $g$ is a left homotopy inverse for $f$. But how can I produce a right homotopy inverse?

Update 2: (In the following I use $\simeq$ for pointed homotopy equivalence.) Conley's specific situation deals with the following portion of the Barratt-Puppe sequence:

$$A \xrightarrow{g} X \hookrightarrow C_g \xrightarrow{f} SA \xrightarrow{Sg} SX.$$

Here $C_g$ is the mapping cone of $g$; $SA, SX$ are reduced suspensions; $Sg$ is the suspension of the map $g$. Also, $$f = C_g \to C_g/X \xrightarrow{\simeq}SA$$ is the composition $C_g \to C_g/X$ with the homotopy equivalence $SA \to C_g/X$ induced by the inclusion of $(A\times I) \sqcup X$ followed by the quotient map to $C_g$ and then the collapsing of the subspace $X$ of $C_g$ (see Bredon p.447, Cor. 5.5).

In Conley's situation that I referenced, he has $X \simeq *$, i.e., $X$ contractible. But assuming that $A$ is well-pointed, then it follows that $X\hookrightarrow C_g$ is always a cofibration$^\mathbf{1}$; since $X$ is also contractible here, we have that $C_g \to C_g/X$ is a homotopy equivalence (c.f. Bredon p. 445, Thm 4.5). Hence $$f = C_g \xrightarrow{\simeq} C_g/X \xrightarrow{\simeq} SA$$ is a pointed homotopy equivalence as desired.

If this argument is correct, then it seems I may have been mistaken in interpreting Conley -- he may have only been asserting $f$ to be a homotopy equivalence via reasoning similar to mine here, rather than asserting anything about general coexact sequences.

Footnotes:

$\mathbf{1}$. (So far I have found this asserted on the web, e.g. here, but not proved, so here is my own proof attempt.) Let $I = [0,1]$, and in what follows place $\times$ before $\cup$ and $/$ in the "order of operations". Since $A$ is well-pointed, the proof of Bredon's Thm 1.9 on p. 436 shows that $A \times \partial I \cup \{*\} \times I \hookrightarrow A\times I$ is a cofibration. Hence the converse part of Bredon's Thm 1.5 on pp.431-432$^\mathbf{2}$ implies that there is a function $\phi_0:A\times I \to [0,1]$ and a neighborhood $U_0 \subset A\times I$ of $A \times \partial I \cup \{*\}$ satisfying (1-3) of Bredon's Thm 1.5 on p. 432. It follows that $\phi_0$ descends to a map $\phi_1$ on the reduced cone

$$CA := A\times I/(A\times \{1\} \cup \{*\}\times I)$$

such that the image $U_1$ of $U_0$ in the quotient $CA$ and $\phi_1$ satisfy the hypotheses of Bredon's Thm 1.5 on p.432. Here $\phi_1^{-1}(0)$ is the base of $CA$ (note that the image of the "crease" $\{*\} \times I$ through the quotient map is included in this base). By the universal property of the quotient topology (applied to the quotient $CA \sqcup X \to C_g$), $\phi_1$ extends to a continuous map $\phi_2$ on $C_g$ with $\phi_2|_X = 0$. Additionally, the neighborhood $U_2\subset C_g$ obtained through the union of the images of $X$ and $U_1$ in $C_g$ is such that $\phi_2, U_2$ is a pair satisfying the hypotheses of Bredon's Thm 1.5 on p. 432. Hence the inclusion $X\hookrightarrow C_g$ of $X$ into the reduced mapping cone $C_g$ is a cofibration.

$\mathbf{2}$. Alternatively, see part (i) of the Theorem at the bottom of p.45 of May’s revised “A Concise Course in Algebraic Topology” which is freely available here.

Matthew Kvalheim
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    I don't know if this works, but I'd try looking at the exact sequence of pointed sets with $Y=A$. – Angina Seng Jan 25 '19 at 08:10
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    Question: the definition of coexact involves three spaces but your question involves four spaces. We do know that $f:A \to B$ is a homotopy equivalence if and only if $f^*; [C,B] \to [C,A]$ is a bijection for all pointed spaces $C$, – Ronnie Brown Jan 25 '19 at 15:49
  • @RonnieBrown: By coexactness of a longer sequence I mean that each short subsequence consisting of two maps is coexact. I've edited the question to clarify. What you say we do know sounds very close to what I would like to know, but I'm wondering whether you have a typo. Did you mean to say $[B,C]$ and $[A,C]$ rather than $[C,B]$ and $[C,A]$? – Matthew Kvalheim Jan 25 '19 at 17:40
  • @LordSharktheUnknown nice idea -- I managed to prove that $f$ has a left homotopy inverse with that approach. But I haven't figured out how to show that $f$ has a right homotopy inverse. – Matthew Kvalheim Jan 25 '19 at 17:40
  • @Max presumably using the map $g \circ f : B\to B$ (with $g$ the left homotopy inverse)? I tried this, but haven't figured out how to make it work. – Matthew Kvalheim Jan 25 '19 at 17:49
  • For clarity, and to agree with your question, I could perhaps have replaced $C$ by $Y$. What I wrote was a special case of a basic fact on representable functors (should be in any book on category theory) - it is a very useful fact, even if the proof is "trivial". The bijection condition for pointed sets seems much stronger than the condition you give from Bredon's book. – Ronnie Brown Jan 25 '19 at 18:43
  • @RonnieBrown: interesting - I am not familiar with representable functors but I will have a look. Thank you. And I think I see -- am I correct that you're saying that the condition from Bredon's book does not appear to imply the bijection condition you mentioned? – Matthew Kvalheim Jan 25 '19 at 19:00
  • The Bredon condition is modelled on groups. But for the exact sequences at the bottom end where based sets are involved you really need to look at the exact sequence(s) of a fibration of groupoids - which involves operations as well as sets with base point. See my topology book. – Ronnie Brown Jan 25 '19 at 22:14

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The following is a nonelementary solution for the case where $A,B$ are simply connected CW-complexes. This is not at all elementary and I expect there should be a way easier solution (without the condition that $A,B$ be simply connected CW-complexes) if it's true. In particular, it uses the natural isomorphism $H^n(-;G) \cong [-;K(G,n)]$

Unwrapping the definition of coexactness, we get that the coexactness of this sequence amounts to two facts :

(i) For all $k:B\to Y$, $k\circ f\simeq * \implies k\simeq *$

(ii) For all $g:A\to Y,$ there exists $\alpha$ with $g\simeq \alpha\circ f$.

Taking $Y=K(G,n)$ for an abelian group $G$ and an integer $n$, knowing that the $0$ element of $H^n(X,G)$ is represented by the nullhomotopic maps $X\to K(G,n)$, we get that by (i) the map $H^n(B,G)\to H^n(A,G)$ has trivial kernel (is injective), and by (ii) is surjective, hence is an iso.

Now because $A,B$ are simply connected CW-complexes, this implies that this is a weak-equivalence, hence a homotopy equivalence.

I had to use the isomorphism I mentioned at the beginning to get from a "local injectivity" (condition (i)) to actual injectivity. Note that, as Ronnie Brown points out in the comments, for $f$ to be a homotopy equivalence, $[B,Y]\to [A,Y]$ has to be bijective for all $Y$, which seems far stronger than conditions (i) and (ii) (specifically, injectivity seems much stronger than condition (i), while condition (ii) is exactly surjectivity)

I'll probably delete this answer once someone comes up with an elementary solution (if it exists, I'll probably leave it there if someone finds a counterexample because it will show that under more conditions the statement remains true)

Maxime Ramzi
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  • Even if someone posts another solution, FWIW reading this was helpful for me. Here's a question: why is an isomorphism on all cohomology groups assuming (simply connected CW-complexes) sufficient to conclude weak homotopy equivalence? My feeling is this must be a standard theorem I don't know. – Matthew Kvalheim Jan 25 '19 at 19:13
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    Yes, it's a standard theorem. See at "What homology isomorphism tells us" https://topospaces.subwiki.org/wiki/Homology_isomorphism_of_topological_spaces It's stated there for homology, but the same statement holds for cohomology, because of e.g. https://math.stackexchange.com/questions/1782321/isomorphism-on-cohomology-implies-isomorphism-on-homology (see the answer; I used all groups, but actually since the complex in question consists of free abelian groups, $K(\mathbb{Z},n)= (S^1)^n$ suffices ) – Maxime Ramzi Jan 25 '19 at 19:56
  • Thanks for the link and comments. By the way, I wrote an "update 2" in my post. If my reasoning is correct, then it seems I might have been mistaken: Conley may have been asserting only something about a very specific coexact sequence rather than general coexact sequences. In any case I would be happy to accept your answer since (at least so far) it comes closest to answering the specific question I originally asked (rather than address the specific situation that I secretly wanted). But if you can spare a minute, I would very much appreciate any feedback on my "update 2". – Matthew Kvalheim Jan 25 '19 at 21:03
  • About your update 2, the fact that $X\simeq $ does not imply that the quotient is a homeomorphism. In fact the quotient is not injective unless $X=$, so it's not a homeomorphism. But under nice hypotheses a quotient by a contractible subspace is a homotopy equivalence (it's the case when the inclusion of the subsace is a cofibration). It will be the case I guess if $X,A$ are CW-complexes – Maxime Ramzi Jan 25 '19 at 21:15
  • Wow, great point -- thank you very much for pointing that out! I think that $X \hookrightarrow C_g$ is always a cofibration, where $X$ is the base of the mapping cone of a map $A \xrightarrow{g} X$. (Do you agree?) I edited my answer and also added a footnote supporting this claim that I just made. – Matthew Kvalheim Jan 25 '19 at 21:55
  • I can't really follow your proof that it is a cofibration (I don't know the theorem you're using, nor do I have the book), it may be true (I think I found some references online stating just that, but I'm not sure because they seemed to be referring to the unpointed mapping cone - is that what you're referring to as well ?) – Maxime Ramzi Jan 25 '19 at 22:38
  • Ok I just got my hands on said theorem. So $\phi$ sends the vertex so $1$ and $U= C_g\setminus$ the vertex ? – Maxime Ramzi Jan 25 '19 at 22:52
  • There is a similar (but possibly more general, I think) version of that theorem at the bottom of p.45 of Peter May’s “A Concise Course in Algebraic Topology” which is freely available in May’s website. Good point about pointed vs unpointed; I had the pointed mapping cone in mind, but I realize now that Bredon’s (and May’s) Theorem is for the unpointed setting. However Bredon states on p. 435 that the result goes through in the pointed setting as well, and according to him it is just a matter of keeping track of the basepoint. I haven’t checked in detail but I personally feel OK accepting that – Matthew Kvalheim Jan 25 '19 at 23:57
  • Yes exactly, $\phi$ sending the vertex to $1$ and $U= C_g\setminus$ the vertex is exactly what I had in mind. – Matthew Kvalheim Jan 25 '19 at 23:59
  • But in the pointed setting, $X$ is not a retract of $U$ because $U$ is missing the basepoint of $X$ (which is "the vertex"). I do believe the result is true in the unpointed case though – Maxime Ramzi Jan 26 '19 at 08:36
  • I think that the reduced mapping cone (pointed setting) of a map $A\to X$ is obtained from the unreduced mapping cylinder by collapsing both $A$ and ${*}\times [0,1]$ to a point, whereas the unreduced mapping cone is obtained by collapsing only $A$ to a point. I.e. the reduced mapping cone also collapses a “crease” in the cone. So I think the deformation retract I described (on the unreduced cone) would induce a pointed deformation retract on the reduced cone. Or am I missing something? – Matthew Kvalheim Jan 26 '19 at 16:57
  • What I'm saying is tha if $U$ is this open set, then since the base point of $X$ is (identified to) the removed vertex, there is no (I mean no obvious) inclusion $X\to U$ – Maxime Ramzi Jan 26 '19 at 17:00
  • Excellent point again -- you've caught a lot my mistakes I likely wouldn't have otherwise caught, so thanks a lot. I think my old argument was indeed wrong for this reason. I searched the web and apparently one needs to assume that $A$ is well-pointed for the inclusion of the base into the reduced mapping cone $C_g$ (with $g:A\to X$) to be a cofibration. I found this asserted on the web but couldn't find a proof, so I deleted the incorrect (unpointed) proof from my update 2 and wrote a new proof using this well-pointedness assumption. – Matthew Kvalheim Jan 26 '19 at 21:01
  • I don't understand the point 2. of your edit, because there still doesn't seem to be a map $C_g\to I$ defined everywhere (or at least I don't see it), precisely because the vertex is at the same time at the $1$-part of the cone, and at the $0$-part. About the reference you gave though (so your point 1.), in the document the map $Y\to C_(f)$ is automatically a cofibration as the pushout of $X\to C_(X)$ which is a cofibration, so it suffices to prove this second point; perhaps it's easier ? (I haven't read your proof yet, I'll have to look at all the references you give) – Maxime Ramzi Jan 26 '19 at 21:10
  • Ah, I see. I think the proof you suggest is roughly equivalent to what I wrote, but is much shorter. I think it is roughly equivalent because I use part of the proof of Bredon's Thm 1.9, and this result contains the statement that $X\to C_*(X)$ is a cofibration. Additionally where I wrote something about "universal property of the quotient topology", I could have just appealed to the universal property of the pushout (which I think the quotient topology universal property must be used to prove). – Matthew Kvalheim Jan 26 '19 at 22:16