If $U : \mathbf{C}/X \to \mathbf{C}$ is the forgetful functor and $\mathbf{C}$ has binary products then I want to find the right adjoint of $U$.
Here is what I have so far:
I will define $F : \mathbf{C} \to \mathbf{C}/X$ by $F(C) = \pi_1 : X \times C \to X$ and I want to show that $F$ is the right adjoint of $U$.
I know that a characterization of $F$ being right adjoint to $U$ is that for any $f : C \to X$ in $\mathbf{C}/X$ and any $C^\prime$ in $\mathbf{C}$ there is an isomorphism $\phi : \text{Hom}_\mathbf{C}(U(f : C \to X),C^\prime) \to \text{Hom}_{\mathbf{C}/X}(f : C \to X, F(C^\prime))$ that is natural in both $f : C \to X$ and $C^\prime$. In this case $\phi$ is equivalent to $\phi : \text{Hom}_\mathbf{C}(C,C^\prime) \to \text{Hom}_{\mathbf{C}/X}(f : C \to X, \pi_1 : X \times C^\prime)$.
For any $g \in \text{Hom}_\mathbf{C}(C,C^\prime)$ there is a unique $\langle f,g \rangle : C \to X \times C^\prime$ such $\pi_1 \circ \langle f,g \rangle = f$ and $\pi_2 \circ \langle f,g \rangle = g$ by the UMP for products. Define $\phi(g) = \langle f,g \rangle$. Then $\phi$ is surjective since for each $h \in \text{Hom}_{\mathbf{C}/X}(f : C \to X, \pi_1 : X \times C^\prime)$ we have that $\pi_2 \circ h : C \to C^\prime$ is in $\text{Hom}_\mathbf{C}(C,C^\prime)$ and is mapped to $h$ by $\phi$. Also, $\phi$ is injective since $\phi(g) = \phi(h)$ implies that $\langle f,g \rangle = \langle f,h \rangle$ and it follows that $g = \pi_2 \circ \langle f,g \rangle = \pi_2 \circ \langle f,h \rangle = h$. Therefore, $\phi$ is an isomorphism.
I now need to show that $\phi$ is natural in $f : C \to X$ and $C^\prime$. I am not sure what this means. I know what a natural isomorphism is, but not what it means for an isomorphism to be natural in an object. What conditions must hold for $\phi$ to be natural in $f$ and $C^\prime$? Any help is much appreciated.
Best Answer
This might be an extended comment rather than an answer.
The idea is not being natural in an object but on an input. So when you fix an input of $\phi : \text{Hom}_\mathbf{C}(U(f : C \to X),C^\prime) \to \text{Hom}_{\mathbf{C}/X}(f : C \to X, F(C^\prime))$ what you get is a natural. In other words, the map $\phi_1 : \text{Hom}_\mathbf{C}(U(f : C \to X),-) \to \text{Hom}_{\mathbf{C}/X}(f : C \to X, F(-))$ is natural (whenever you have a morphism $C'\to C''$ you get the usual commutative diagram between $\phi_1(C')$ and $\phi_2(C'')$). And similarly on the second input, i.e. the map $\phi_2 : \text{Hom}_\mathbf{C}(U(-),C') \to \text{Hom}_{\mathbf{C}/X}(-, F(C'))$ is natural.
Indeed you can do both inputs at once if you consider $\text{Hom}$ to be a functor from the product category (also called bifunctor) so that a morphism between $(f:C\to X, C')\to (f'':C''\to X, C''')$ is just a pair of morphisms, each component corresponding to a component of these pairs. This way you get that $\phi : \text{Hom}_\mathbf{C}(U(-),-) \to \text{Hom}_{\mathbf{C}/X}(-, F(-))$ is natural.