# What is exponentiation in the slice category?

I'm looking for an explicit construction, however involved, of the exponential objects in the slice categories of a topos. If we have a (elementary) topos $$\mathbf{C}$$ (cartesian closed+subobject classifier) and an object $$a\in\mathbf{C}$$, what is the exponential $$g^f$$ of the $$\mathbf{C}/a$$-objects $$(f:x\to a)$$ and $$(g:y\to a)$$ ?

The exponential object is a $$\mathbf{C}$$-arrow $$g^f:E\to a$$ but what is the object $$E$$ ? Maybe $$y^x$$ in $$\mathbf{C}$$ ? I'm trying to change diagrams of $$\mathbf{C}/a$$ into diagrams of $$\mathbf{C}$$ (for example from a product diagram you get a pullback diagram and vice versa in some cases) but it hasn't worked out yet.

What should I be looking for to construct the object $$E$$, the arrow $$g^f:E\to a$$ and the evaluation map $$ev:g^f\times f\to g$$ (arrow and product in $$\mathbf{C}/a$$) ?

I know could get the result easier with adjoint functors but I want the explicit step by step construction

Thanks!

• @MaliceVidrine Thanks. The original question is in the context of topos theory. I'll edit the question so it admits an answer Apr 16, 2020 at 20:34
• You can start from the case when C is a presheaf topos! Apr 16, 2020 at 21:38
• I know of only one construction (from Johnstone's Topos Theory, theorem 1.42) that doesn't go through the internal language or just directly use the existence of a right adjoint to pullback functors. But I don't understand Johnstone's proof, so I'm not sure how helpful knowing the construction is without the proof that it works. Apr 17, 2020 at 1:08

I believe I've successfully decoded the proof from Johnstone I mention in the comments. While I'm writing this much more explicitly than he did, I do recommend drawing out the details in a notebook to fill in the details I gloss over in the first couple of steps. They're the easier steps, though.

To start, we need the notion of partial map classifiers, which always exist in a topos. Given $$f:A\to X$$, we define the map $$\theta:A\times X\to\tilde X$$ as the map that classifies the partial morphism $$A\times X\overset{\langle 1_A,f\rangle}{\leftarrowtail}A\overset{f}{\to}X$$. Then for any $$g:B\to X$$, the exponential $$g^f$$ is given by the pullback of $$\tilde{g}^A:\tilde{B}^A\to\tilde{X}^A$$ along $$\bar{\theta}:X\to\tilde{X}^A$$, which I'll denote $$g^f:E\to X$$.

Why should this be the exponential? If you were to look at what $$\bar{\theta}$$ does in the category $$\mathbf{Set}$$, it takes every element of $$x$$ to the partial map $$\gamma_x$$ such that $$\gamma_x(a)=x$$ if $$x=f(a)$$, and is undefined otherwise---that is, $$\gamma_x$$ is the restriction of $$f$$ to $$f^{-1}(\{x\})$$. The object part of the pullback is then the set of $$\langle x,m\rangle$$ where $$m$$ is a partial map from $$A\to B$$ such that $$g\circ m=\gamma_x$$; this says all at once that the domain of $$m$$ is exactly $$f^{-1}(\{x\})$$ and that its range is within $$g^{-1}(\{x\})$$. As it turns out, this is one of the surprisingly common cases where the set theoretic analogy pays off in an arbitrary topos.

So suppose we have $$\require{AMScd}\begin{CD} T @>z>> E \\ @VhVV @VVg^fV \\ X @= X \end{CD}$$ Some diagram shuffling and composition with evaluation maps gives us a morphism over $$\tilde X$$ $$\require{AMScd}\begin{CD}A\times T @>ev\circ 1_A\times(p_1 z)>> \tilde{B} \\ @V{\theta\circ 1_A\times h}VV @VV\tilde g V \\ \tilde X @= \tilde X \end{CD}.\qquad (1)$$

At this point we need to note that the left hand side of the previous square corresponds to the partial morphism given by the top and left edges of $$\require{AMScd}\begin{CD} A\times_X T @>p_1>> A @>f>> X \\ @VVV @VV{\langle 1_A,f\rangle}V \\ A\times T @>>{1_A\times h}> A\times X \end{CD}\qquad (2)$$ where the square is a pullback and $$A\times_X T$$ the object part of that pullback because $$\require{AMScd}\begin{CD} A\times T @>1_A\times h>> A\times X \\ @V\pi_2VV @VV\pi_2V \\ T @>>h> X \end{CD}$$ is a pullback so we can apply the pullback lemma by pasting this onto the square in (2).

Because (1) commutes, this means the diagram $$\require{AMScd}\begin{CD} A\times T @<<< A\times_X T \\ @Vev\circ 1_A\times(p_1 z)VV @| \\ \tilde{B} @. A\times_X T \\ @V\tilde{g}VV @VVf p_1V \\ \tilde X @<<< X \end{CD}$$ is a pullback. Since $$\require{AMScd}\begin{CD} \tilde{B} @<<< B \\ @V\tilde{g}VV @VVgV \\ \tilde{X} @<<< X \end{CD}$$ is a pullback, its universal property gets us a unique $$\require{AMScd}\begin{CD} A\times _X T @>>> B \\ @Vfp_1VV @VVgV \\ X @= X \end{CD}$$ as desired. Each of these steps is also reversible, so this establishes the desired universal property.

• Thank you! A lot! Apr 18, 2020 at 14:14
• No problem! It was good to finally understand this better myself! Apr 18, 2020 at 20:02