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I am currently reading Lemma 9.11 on Switzer, in the chapter dedicated to Brown representability theorem. However I am stuck on a few points of the proof.

Let $F$ a contravariant functor from pointed CW complexes up to homotopy to pointed sets. Let $Y$ be a pointed CW complex, $u \in F(Y)$ and $(X, A, x_0)$ a CW-pair. Let $g : (A, x_0) \to (Y, y_0)$ a cellular map. Let $T =( I^{+} \wedge A) \vee X \vee Y$, by identifying $[0,a] \in I^{+} \wedge A$ with $a \in X$ and $[1,a] \in I^{+} \wedge A$ with $g(a) \in Y$. Let $A_1 =( [0, 1/2]^{+} \wedge A) \cup X$ and $A_2 =( [1/2, 1]^{+} \wedge A) \cup Y$. Now $A_1 \cup A_2 = T, A_1 \cap A_2$ is isomorphic to $A$. Until this point it is all clear. What I don’t understand is the following:

There is a strong deformation retraction $f: A_1 \to X$ and $Y$ is a strong deformation retract of $A_2$. Thus there are $\bar{v} \in F(A_1)$ with $\bar{v}|_{X} = v$ and a $\bar{u} \in F(A_2)$ with $\bar{u}|_{Y}= u$. Why is that?

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  • $\begingroup$ Probably $T = A_1 {\color{red} \cup} A_2$, not the intersection? Can you draw a picture of $T, A_1, A_2$? Also, who is $v$ (an element of $FX$?), and what about $(-)|_X, (-)|_Y$? An element $\bar u$ such that $Fr(\bar u)=u\in FY$, for some $r : A_2\to Y$ comes form the fact that $Y$ is a sdr of $A_2$, so you get a surjective map $FA_2 \to FY$ (every functor sends retracts to retracts, which are absolute colimits). $\endgroup$
    – fosco
    Commented Feb 9, 2021 at 10:11
  • $\begingroup$ I managed to do̶w̶n̶l̶ find a copy of Switzer and the element $u$ isn't just any element, it is a universal element; this means first of all that $F$ is contra variant (as it must be, if in the end it will represent cohomology), and that there is an isomorphism $\pi_n(Y)\cong F(S^n)$ for all $n\ge 0$, induced by evaluating on $u$ as per def. 9.6; also, $F$ is probably required to satisfy some form of cohomology axiom (preserving coproducts and Mayer-Vietoris sequences)? The above isomorphism makes sense only when $F(S^n)$ is naturally a(n abelian) group... $\endgroup$
    – fosco
    Commented Feb 9, 2021 at 10:46
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    $\begingroup$ The superficial and one-line answer is that cohomology theories are contravariant functors, and there's no such thing as a BRT for homology theories. The reason why there is no such BRT is probaby the second thing too large to expand for this narrow margin... a form of BRT says that a functor $F : \mathcal C^{op} \to Set$ defined on a category generated by a strong separating family made of cogroup objects is representable if and only if it preserves products and it is a Mayer-Vietoris functor. $\endgroup$
    – fosco
    Commented Feb 12, 2021 at 10:45
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    $\begingroup$ Let me end with the punchline: by Yoneda lemma, a presheaf $F : \mathcal C^{op} \to Set$ is representable if and only if its category of elements has a terminal object, and all the pain one has to undergo to prove BRT is meant to construct such object. This is attained using the assumption that $\mathcal C$ is generated under colimits by the separating family above. Also, this is how Strom tries to convince you that there is no conceptual reason for homology to exist. $\endgroup$
    – fosco
    Commented Feb 12, 2021 at 11:08
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    $\begingroup$ So it seems, but I wouldn't judge this as a particularly deep insight; the reason why representability of $F$ amounts to the existence of a terminal object $T$ in $Elts(F)$ is that terminality in $Elts(F)$ is, word-by-word, the request that there exists a natural isomorphism $F(-) \cong hom(-,T)$. $\endgroup$
    – fosco
    Commented Feb 12, 2021 at 11:23

1 Answer 1

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Your space $T$ is nothing else than the reduced double mapping cylinder of the maps $\iota : A \hookrightarrow X$ and $g : A \to Y$. To see that, note that $I^+ \wedge A = (I \times A)/(I \times \{a_0\})$.

The subspaces $A_1$ and $A_2$ of $T$ are copies of the reduced mapping cylinders of $\iota$ and of $g$. Thus you get the usual strong deformation retractions $f : A_1 \to X$ and $r : A_2 \to Y$. Hence the inclusions $i_1 : X \to A_1$ and $i_2 : Y \to A_2$ are homotopy equivalences and induce bijections $i_1^* : F(A_1) \to F(X)$ and $i_2^* : F(A_2) \to F(Y)$.

Now consider $v \in F(X)$ and $u \in F(Y)$.

  1. Let $\bar v = (i_1^*)^{-1}(v) \in F(A_1)$. Then by definition $\bar v \mid_X = i^*_1(\bar v) = i^*_1( (i_1^*)^{-1}(v)) = v$.

  2. Let $\bar u = (i_2^*)^{-1}(u) \in F(A_2)$. Then $\bar u \mid_Y = u$.

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  • $\begingroup$ Sorry to bother again, I was rereading your answer: could you please give me reference for the implication “$i_1$ is an homotopy equivalence, therefore $i_1^{*}$ is a bijection?” Is it because homotopy equivalences are isomorphisms in the domain category, therefore they are preserved by $F$? In that case, is $F$ fully faithful? $\endgroup$
    – cip
    Commented Feb 18, 2021 at 9:30
  • $\begingroup$ @cip Yes, each functor preserves isomorphisms; it is irrelevant whether $F$ is full or faithful. Also note that the $i_k$ are no isomorphisms, but induce isomorphisms in the homotopy category on which $F$ lives. $\endgroup$
    – Paul Frost
    Commented Feb 18, 2021 at 9:46
  • $\begingroup$ Yes, I realised I had not specified in my question that the domain category is that of pointed CW-complexes up to homotopy. Thanks a lot! :) $\endgroup$
    – cip
    Commented Feb 18, 2021 at 10:10

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