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What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}} \newcommand{\op}[1]{#1^{\mathrm{op}}} \newcommand{\Set}{\mathrm{Set}}$

In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's".

The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category.

More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by $$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$ where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.

What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}} \newcommand{\op}[1]{#1^{\mathrm{op}}} \newcommand{\Set}{\mathrm{Set}}$

In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's".

The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category.

More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by $$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$ where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.

What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}} \newcommand{\op}[1]{#1^{\mathrm{op}}} \newcommand{\Set}{\mathrm{Set}}$

In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's".

The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category.

More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by $$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$ where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.

What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}} \newcommand{\op}[1]{#1^{\mathrm{op}}} \newcommand{\Set}{\mathrm{Set}}$

In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's".

The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category.

More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by $$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$ where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.

What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}} \newcommand{\op}[1]{#1^{\mathrm{op}}} \newcommand{\Set}{\mathrm{Set}}$

In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's".

The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category.

More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by $$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$ where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.

What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}} \newcommand{\op}[1]{#1^{\mathrm{op}}} \newcommand{\Set}{\mathrm{Set}}$

In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's".

The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category.

More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by $$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$ where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.