It is common for the structure of objects in a category is corepresentable, and so the formula $$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$ is a formula for the right adjoint.

As an example of how this works, in the category of groups, you have the fact that $$ |G| \cong \hom(\mathbb{Z}, G) $$ where $|G|$ means the underlying set, and that the group operation is given by the natural transformation $$ \hom(\mathbb{Z}, G) \times \hom(\mathbb{Z}, G) \cong \hom(F_2, G) \to \hom(\mathbb{Z}, G) $$ induced by the map $\mathbb{Z} \to F_2$ sending $1 \mapsto xy$, where $F_2$ is the free group on two elements $x$ and $y$. (it is the coproduct of $\mathbb{Z}$ with itself, thus the first isomorphism)

So, if $f^\bullet$ is a group-valued functor, you can compute it as follows:

• The underlying set of $f^\bullet X$ is given by $\hom(f_\bullet \mathbb{Z}, X)$
• The group operation is $\hom(f_\bullet \mathbb{Z}, X) \times \hom(f_\bullet \mathbb{Z}, X) \cong \hom(f_\bullet F_2, X) \to \hom(f_\bullet \mathbb{Z}, X)$

and this information suffices to determine the group; the identity and inverse can be given directly by similar formulas if desired.

It is common for the structure of objects in a category to be corepresentable, and so the formula $$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$ is a formula for the right adjoint.

As an example of how this works, in the category of groups, you have the fact that $$ |G| \cong \hom(\mathbb{Z}, G) $$ where $|G|$ means the underlying set, and that the group operation is given by the natural transformation $$ \hom(\mathbb{Z}, G) \times \hom(\mathbb{Z}, G) \cong \hom(F_2, G) \to \hom(\mathbb{Z}, G) $$ induced by the map $\mathbb{Z} \to F_2$ sending $1 \mapsto xy$, where $F_2$ is the free group on two elements $x$ and $y$. (it is the coproduct of $\mathbb{Z}$ with itself, thus the first isomorphism)

So, if $f^\bullet$ is a group-valued functor, you can compute it as follows:

• The underlying set of $f^\bullet X$ is given by $\hom(f_\bullet \mathbb{Z}, X)$
• The group operation is $\hom(f_\bullet \mathbb{Z}, X) \times \hom(f_\bullet \mathbb{Z}, X) \cong \hom(f_\bullet F_2, X) \to \hom(f_\bullet \mathbb{Z}, X)$

and this information suffices to determine the group; the identity and inverse can be given directly by similar formulas if desired.

It is very common that the structure of objects in a category is corepresentable, and so the formula $$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$ is a formula for the right adjoint.

As an example of how this works, in the category of groups, you have the fact that $$ |G| \cong \hom(\mathbb{Z}, G) $$ where $|G|$ means the underlying set, and that the group operation is given by the natural transformation $$ \hom(\mathbb{Z}, G) \times \hom(\mathbb{Z}, G) \cong \hom(F_2, G) \to \hom(\mathbb{Z}, G) $$ induced by the map $\mathbb{Z} \to F_2$ sending $1 \mapsto xy$, where $F_2$ is the free group on two elements $x$ and $y$. (it is the coproduct of $\mathbb{Z}$ with itself, thus the first isomorphism)

So, if $f^\bullet$ is a group-valued functor, you can compute it as follows:

• The underlying set of $f^\bullet X$ is given by $\hom(f_\bullet \mathbb{Z}, X)$
• The group operation is $\hom(f_\bullet \mathbb{Z}, X) \times \hom(f_\bullet \mathbb{Z}, X) \cong \hom(f_\bullet F_2, X) \to \hom(f_\bullet \mathbb{Z}, X)$

and this information suffices to determine the group; the identity and inverse can be given directly by similar formulas if desired.

It is common for the structure of objects in a category is corepresentable, and so the formula $$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$ is a formula for the right adjoint.

As an example of how this works, in the category of groups, you have the fact that $$ |G| \cong \hom(\mathbb{Z}, G) $$ where $|G|$ means the underlying set, and that the group operation is given by the natural transformation $$ \hom(\mathbb{Z}, G) \times \hom(\mathbb{Z}, G) \cong \hom(F_2, G) \to \hom(\mathbb{Z}, G) $$ induced by the map $\mathbb{Z} \to F_2$ sending $1 \mapsto xy$, where $F_2$ is the free group on two elements $x$ and $y$. (it is the coproduct of $\mathbb{Z}$ with itself, thus the first isomorphism)

So, if $f^\bullet$ is a group-valued functor, you can compute it as follows:

• The underlying set of $f^\bullet X$ is given by $\hom(f_\bullet \mathbb{Z}, X)$
• The group operation is $\hom(f_\bullet \mathbb{Z}, X) \times \hom(f_\bullet \mathbb{Z}, X) \cong \hom(f_\bullet F_2, X) \to \hom(f_\bullet \mathbb{Z}, X)$

and this information suffices to determine the group; the identity and inverse can be given directly by similar formulas if desired.