Proof that a category is cartesian closed

category-theorytopos-theory

Suppose that a small category $$\mathbf C$$ has all the finite products. Then the category $$\mathbf{Set}^{\mathbf{C}^{\mathrm {op}}}$$ is cartesian closed.

Fix a $$P\in \mathbf{C}^{\mathrm {op}}$$. Then we know how $$-\times P$$ is defined, and from Yoneda's lemma we can find its right adjoint $$-^P$$. In fact, for any $$Q\in \mathbf{C}^{\mathrm {op}}$$, we have that on an object $$C\in \mathbf{C}$$ our functor $$Q^P(C)$$ is defined by $$\operatorname {nat}(H_C,Q^P)$$, and so by $$\operatorname {nat}(H_C\times P,Q)$$. My difficulties are in proving the universal property of the counit $$\epsilon: Q^P\times P\xrightarrow{_.} Q$$, that is defined by $$\epsilon_C:(\theta,z)\mapsto \theta(1_C,z)$$ for any $$C$$.

I must prove that (for $$R\in \mathbf{C}^{\mathrm {op}}$$) given a natural transformation $$\phi:R \times P \xrightarrow{_.} Q$$, there is a unique $$\bar{\phi}$$ such that $$\epsilon\circ (\bar{\phi}\times 1_{P})=\phi$$. So take a natural transformation $$\alpha:R\xrightarrow{_.} Q^P$$ and suppose that it makes the triangular diagram of the counit commutative. Then for any object $$C\in \mathbf{C}$$, study the $$\alpha_C$$; we must define $$\alpha_C$$ on the elements $$x\in RC$$. That means, we must define the natural transformation $$\alpha_C(x):H_C\times P\xrightarrow{_.} Q$$ at every level, i.e. define any $$[\alpha_C(x)]_D$$ for $$D\in \mathbf{C}$$. Since I know that $$[\alpha_C(x)]_C(1_C,z)$$ is equal to $$\phi(x,z)$$, I presume that the kind of argument that I should use involves Yoneda / universal elements. However in general I must define $$[\alpha_C(x)]_D:H_C(D)\times PD\to QD$$: starting from $$1_C$$ I can surely catch any $$f:D\to C$$, but I don't know how to catch any $$y\in PD$$ starting from $$z\in PC$$ (using $$Pf$$?). I hope I explained well the problem, thanks for any clarify

The point is that when you map $$H_D \times P$$ to $$H_C \times P$$ via $$Hf$$, $$\operatorname{1_D}$$ goes to $$f$$ and $$z$$ is left intact.

What you know is that $$\alpha: R\xrightarrow{} Q^P$$ will be a natural transformation, so for $$f: D \to C$$ this has to commute: $$\require{AMScd}$$ $$\begin{CD} R(C) @>{\alpha_C}>> \{H_C \times P \to Q\} \\ @VRfVV @VVprecomposition \>with\> Hf \times idV \\ R(D) @>{\alpha_D}>> \{H_D \times P \to Q\}. \end{CD}$$

This means that this commutes: $$\begin{CD} H_D \times P @>{\alpha_D(Rf(x))}>> Q \\ @VHf \times idVV @VV=V \\ H_C \times P @>{\alpha_C(x)}>> Q. \end{CD}$$

In particular, plugging in $$(\operatorname{1_D}, z) \in H_D(D) \times P(D)$$ gives $$\alpha_C(x)(f,z) = \alpha_D(Rf(x))((\operatorname{1_D}, z)) = \phi(Rf(x),z)$$.