# Adjoint(s) to the forgetful functor $U:A/\mathbf{C}\to \mathbf{C}$.

I am preparing for my exam in Category Theory, and came across the following exercise in an old exam. Let $$\mathbf{C}$$ a category with finite coproducts. For a fixed object $$A$$, consider the coslice category consisting of objects $$f:A\to C$$. Morphisms are $$\alpha:C\to D$$ making the triangle commute. We have to determine whether the forgetful functor $$U$$ has a left or / and right adjoint.

An (rather unfounded) approach I had in mind for the right adjoint was the functor $$F$$ which maps an object $$C$$ to $$i_A:A\to A\sqcup C$$, where $$i_A$$ denotes the inclusion map. A morphism $$\alpha:C\to D$$ is then mapped to the unique $$u:A\sqcup C\to A \sqcup D$$ which arises when considering the maps $$i_A:A\to A\sqcup D$$ and $$i_D\circ f:C\to A\sqcup D$$, by the universal property of the coproduct. Since this functor does not preserve the terminal object it can't be the left adjoint. To show it is indeed a right adjoint we need to show the following isomorphism of Hom sets:

$$\hom_{\mathbf{C}}(D,U(f:A\to C))\cong \hom_{A/\mathbf{C}}(i_A:A\to A\sqcup D,f:A\to C)$$

However, I failed to show this and do not have an alternative idea so far. Neither do I have an idea for a possible left adjoint, if it exists.

Any kind of help is welcome!

• I corrected a small typo and added some notation, maybe now it is clearer what you have to do ? Jan 17, 2020 at 15:05
• Also the functor you describe $[D \mapsto (i_A : A \to A \sqcup D)]$ is a left adjoint. Jan 17, 2020 at 15:52
• But It does not preserve the terminal object, does it?
– EBP
Jan 17, 2020 at 16:07
• Sorry I didn't correct everything, so the functor $[D \mapsto (i_A : A \to A \sqcup D)]$ preserves the initial object, indeed $(i_A : A \to A \sqcup 0) = id_A$, and $id_A$ is the initial object of $A/ \mathbf(C)$. Jan 17, 2020 at 16:08
• a left adjoint has to preserve colimits, and so the initial object since it is the empty colimit, but there is nothing to be said with limits, and your category $\mathbf{C}$ maybe has not a final object. Jan 17, 2020 at 16:13

For the fact that it admits a left adjoint your (not unfounded at all) discussion gives you the awnser (the only problem is you were trying the wrong side) : $$\text{Hom}_{\mathbf{C}}(D, U(f:A \to C)) \cong \text{Hom}_{A/\mathbf{C}}(i_A : A \to A \sqcup D, f:A\to C).$$ Indeed having a map $$g : D \to C$$ will give, by universal property of $$A\sqcup D$$ and the given data $$f:A \to C$$, a map $$\overline g : A\sqcup D \to C$$ that verifies $$\overline g \circ i_A = f$$, i.e. $$\overline g$$ is a morphism in $$A/\mathbf C$$ from $$i_A : A \to A\sqcup D$$ to $$f:A\to C$$.
For the right adjoint part, if $$U$$ admits a right adjoint, this would mean that $$U$$ is left adjoint, and so it should at least preserve the intial object, but the initial object of $$A/\mathbf{C}$$ is $$id_A : A \to A$$, and $$U(id_A)= A$$ which is not a priori the intial object of $$\mathbf{C}.$$