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.


$\mathbf{1}$. (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.

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    $\begingroup$ I don't know if this works, but I'd try looking at the exact sequence of pointed sets with $Y=A$. $\endgroup$ – Angina Seng Jan 25 '19 at 8:10
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    $\begingroup$ 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 $C$, $\endgroup$ – Ronnie Brown Jan 25 '19 at 15:49
  • $\begingroup$ @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 $[C,A]$? $\endgroup$ – Matthew Kvalheim Jan 25 '19 at 17:40
  • $\begingroup$ @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. $\endgroup$ – Matthew Kvalheim Jan 25 '19 at 17:40
  • $\begingroup$ @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. $\endgroup$ – Matthew Kvalheim Jan 25 '19 at 17:49

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)

  • $\begingroup$ 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. $\endgroup$ – Matthew Kvalheim Jan 25 '19 at 19:13
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    $\begingroup$ Yes, it's a standard theorem. See at "What homology isomorphism tells us" topospaces.subwiki.org/wiki/… It's stated there for homology, but the same statement holds for cohomology, because of e.g. math.stackexchange.com/questions/1782321/… (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 ) $\endgroup$ – Maxime Ramzi Jan 25 '19 at 19:56
  • $\begingroup$ 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". $\endgroup$ – Matthew Kvalheim Jan 25 '19 at 21:03
  • $\begingroup$ 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 $\endgroup$ – Maxime Ramzi Jan 25 '19 at 21:15
  • $\begingroup$ 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. $\endgroup$ – Matthew Kvalheim Jan 25 '19 at 21:55

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