Right adjoint of forgetful functor from slice category $\mathbf{C}/X$ to $\mathbf{C}$ 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.
 A: 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.
