show/hide this revision's text 3 added 62 characters in body

I was thinking more about this question and found another paper by Heller which offers an answer (unfortunately a negative one). The paper is

Freyd, Peter; Heller, Alex Splitting homotopy idempotents. II. J. Pure Appl. Algebra 89 (1993), no. 1-2, 93–106.

This paper introduces a notion of conjugacy idempotent. It is a triple $(G, g, b)$ consisting of a group $G$, an endomorphism $g \colon G \to G$ and an element $b \in G$ such that for all $x \in G$ we have $g(x) g^2(x) = b^{-1} x g(x) b$. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent $(F, f, a)$. The main result Main Theorem of the paper is says (among other things) that $f$ does not split in the quotient of the category of groups by the conjugacy congruence.

Now $f$ induces an endomorphism $B f \colon B F \to B F$ which is an idempotent in $\mathrm{Ho} \mathrm{Top}$ and it follows (by the Main Lemma of the paper) that it doesn't split. It is then easily concluded that $(B f)_+ \colon (B F)_+ \to (B F)_+$ doesn't split in $\mathrm{Ho} \mathrm{Top}_*$.

The map $(B f)_+$ induces an idempotent of the representable functor $[-, (B F)_+]_*$ which does split since this is a $\mathrm{Set}$ valued functor. Let $H \colon \mathrm{Ho} \mathrm{Top}_*^\mathrm{op} \to \mathrm{Set}$ be the resulting retract of $[-, (B F)_+]_*$. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is not representable since a representation would provide a splitting for $(B f)_+$.

The same argument with $B f$ in place of $(B f)_+$ shows the failure of Brown's Representability in the unbased case.
show/hide this revision's text 2 added 13 characters in body

I was thinking more about this question and found another paper by Heller which offers an answer (unfortunately a negative one). The paper is

Freyd, Peter; Heller, Alex Splitting homotopy idempotents. II. J. Pure Appl. Algebra 89 (1993), no. 1-2, 93–106.

This paper introduces a notion of conjugacy idempotent. It is a triple $(G, g, b)$ consisting of a group $G$, an endomorphism $g \colon G \to G$ and an element $b \in G$ such that for all $x \in G$ we have $g(x) = b^{-1} x b$. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent $(F, f, a)$. The main result of the paper is that $f$ does not split in the quotient of the category of groups by the conjugacy congruence.

Now $f$ induces an endomorphism $B f \colon B F \to B F$ which is an idempotent in $\mathrm{Ho} \mathrm{Top}$ and it follows that it doesn't split. It is then easily concluded that $(B f)_+ \colon (B F)_+ \to (B F)_+$ doesn't split in $\mathrm{Ho} \mathrm{Top}_*$.

The map $(B f)_+$ induces an idempotent of the representable functor $[-, (B F)_+]_*$ which does split since this is a $\mathrm{Set}$ valued functor. Let $H \colon \mathrm{Ho} \mathrm{Top}_* mathrm{Top}_*^\mathrm{op} \to \mathrm{Set}$ be the resulting retract of $[-, (B F)_+]_*$. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is no not representable since a representation would provide a splitting for $(B f)_+$.

The same argument with $B f$ in place of $(B f)_+$ shows the failure of Brown's Representability in the unbased case.
show/hide this revision's text 1

I was thinking more about this question and found another paper by Heller which offers an answer (unfortunately a negative one). The paper is

Freyd, Peter; Heller, Alex Splitting homotopy idempotents. II. J. Pure Appl. Algebra 89 (1993), no. 1-2, 93–106.

This paper introduces a notion of conjugacy idempotent. It is a triple $(G, g, b)$ consisting of a group $G$, an endomorphism $g \colon G \to G$ and an element $b \in G$ such that for all $x \in G$ we have $g(x) = b^{-1} x b$. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent $(F, f, a)$. The main result of the paper is that $f$ does not split in the quotient of the category of groups by the conjugacy congruence.

Now $f$ induces an endomorphism $B f \colon B F \to B F$ which is an idempotent in $\mathrm{Ho} \mathrm{Top}$ and it follows that it doesn't split. It is then easily concluded that $(B f)_+ \colon (B F)_+ \to (B F)_+$ doesn't split in $\mathrm{Ho} \mathrm{Top}_*$.

The map $(B f)_+$ induces an idempotent of the representable functor $[-, (B F)_+]_*$ which does split since this is a $\mathrm{Set}$ valued functor. Let $H \colon \mathrm{Ho} \mathrm{Top}_* \to \mathrm{Set}$ be the resulting retract of $[-, (B F)_+]_*$. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is no representable since a representation would provide a splitting for $(B f)_+$.

The same argument with $B f$ in place of $(B f)_+$ shows the failure of Brown's Representability in the unbased case.