# Strong deformation retraction between CW-complexes, Brown theorem proof

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?

• 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). Commented Feb 9, 2021 at 10:11
• 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... Commented Feb 9, 2021 at 10:46
• 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. Commented Feb 12, 2021 at 10:45
• 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. Commented Feb 12, 2021 at 11:08
• 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 $F(-) \cong hom(-,T)$. Commented Feb 12, 2021 at 11:23

## 1 Answer

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 $$\bar u \mid_Y = u$$.

• 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?
– cip
Commented Feb 18, 2021 at 9:30
• @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. Commented Feb 18, 2021 at 9:46
• 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! :)
– cip
Commented Feb 18, 2021 at 10:10