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There is no full functor from pointed spaces, or its homotopy category, to groups, because there is no full functor from pointed sets to groups. Proof: Let $F$ be such a functor. Denote by $n$ a based set with $n$ elements. Then the group $F(1)$ must be trivial because it has only one endomorphism; the group $F(2)$ must be of order two because it has exactly two endomorphisms; the group $F(3)$ must have have exactly two automorphisms, making it abelian, and on the other hand it must have exactly three morphisms from $F(2)$. Contradiction.

We could ask about the homotopy category of path-connected based spaces (which I suppose is just as off-putting, Martin, as the category of non-empty spaces, but let's go on anyway). This has a full functor from groups, so if it had a full functor to groups then composing we would get a full functor from groups to groups which is not a equivalence of categories. Sounds impossible, but I haven't got an argument.

There is no full* functor from pointed spaces, or its homotopy category, to groups, because there is no full* functor from pointed sets to groups. Proof: Let $F$ be such a functor. Denote by $n$ a based set with $n$ elements. Then the group $F(1)$ must be trivial because it has only one endomorphism; the group $F(2)$ must be of order two because it has exactly two endomorphisms; the group $F(3)$ must have have exactly two automorphisms, making it abelian, and on the other hand it must have exactly three morphisms from $F(2)$. Contradiction.

We could ask about the homotopy category of path-connected based spaces (which I suppose is just as off-putting, Martin, as the category of non-empty spaces, but let's go on anyway). This has a full functor from groups, so if it had a full functor to groups then composing we would get a full functor from groups to groups which is not a equivalence of categories. Sounds impossible, but I haven't got an argument.

EDIT: * fully faithful

There is no full functor from pointed spaces, or its homotopy category, to groups, because there is no full functor from pointed sets to groups. Proof: Let $F$ be such a functor. Denote by $n$ a based set with $n$ elements. Then the group $F(1)$ must be trivial because it has only one endomorphism; the group $F(2)$ must be trivial because it has exactly two endomorphisms; the group $F(3)$ must have have exactly two automorphisms, making it abelian, and on the other hand it must have exactly three morphisms from $F(2)$. Contradiction.

We could ask about the homotopy category of path-connected based spaces (which I suppose is just as off-putting, Martin, as the category of non-empty spaces, but let's go on anyway). This has a full functor from groups, so if it had a full functor to groups then composing we would get a full functor from groups to groups which is not a equivalence of categories. Sounds impossible, but I haven't got an argument.

There is no full functor from pointed spaces, or its homotopy category, to groups, because there is no full functor from pointed sets to groups. Proof: Let $F$ be such a functor. Denote by $n$ a based set with $n$ elements. Then the group $F(1)$ must be trivial because it has only one endomorphism; the group $F(2)$ must be of order two because it has exactly two endomorphisms; the group $F(3)$ must have have exactly two automorphisms, making it abelian, and on the other hand it must have exactly three morphisms from $F(2)$. Contradiction.

We could ask about the homotopy category of path-connected based spaces (which I suppose is just as off-putting, Martin, as the category of non-empty spaces, but let's go on anyway). This has a full functor from groups, so if it had a full functor to groups then composing we would get a full functor from groups to groups which is not a equivalence of categories. Sounds impossible, but I haven't got an argument.

There is no full functor from pointed spaces, or its homotopy category, to groups, because there is no full functor from pointed sets to groups. Proof: Let $F$ be such a functor. Denote by $n$ a based set with $n$ elements. Then the group $F(1)$ must be trivial because it has only one endomorphism; the group $F(2)$ must be trivial because it has exactly two endomorphisms; the group $F(3)$ must have have exactly two automorphisms, making it abelian, and on the other hand it must have exactly two morphisms from $F(2)$. Contradiction.

We could ask about the homotopy category of path-connected based spaces (which I suppose is just as off-putting, Martin, as the category of non-empty spaces, but let's go on anyway). This has a full functor from groups, so if it had a full functor to groups then composing we would get a full functor from groups to groups which is not a equivalence of categories. Sounds impossible, but I haven't got an argument.

There is no full functor from pointed spaces, or its homotopy category, to groups, because there is no full functor from pointed sets to groups. Proof: Let $F$ be such a functor. Denote by $n$ a based set with $n$ elements. Then the group $F(1)$ must be trivial because it has only one endomorphism; the group $F(2)$ must be trivial because it has exactly two endomorphisms; the group $F(3)$ must have have exactly two automorphisms, making it abelian, and on the other hand it must have exactly three morphisms from $F(2)$. Contradiction.

We could ask about the homotopy category of path-connected based spaces (which I suppose is just as off-putting, Martin, as the category of non-empty spaces, but let's go on anyway). This has a full functor from groups, so if it had a full functor to groups then composing we would get a full functor from groups to groups which is not a equivalence of categories. Sounds impossible, but I haven't got an argument.