I'm reading Emily Riehl's Category Theory in Context and pondering Exercise 1.2.i, where the author writes "Defining $\mathcal{C}/c$ to be $(c/\mathcal{C}^{op})^{op}$," in an effort to exhibit duality between the slice and co-slice categories. If you allow yourself complete freedom to just "reverse all the arrows" in the underlying commutative diagrams of the underlying category $\mathcal{C}$, this makes sense, but I get stuck when I try to step through the construction.
So, we start with $\mathcal{C}$, and we have objects $x,y,c$ and arrows $f:c\rightarrow x$, $g:c\rightarrow y$ and $h:x\rightarrow y$ such that $g=hf$.
Moving to $\mathcal{C}^{op}$, we have the same objects $x,y,c$, and now we have arrows that are the reverse of all the arrows in $\mathcal{C}$. But, the construction of $c/\mathcal{C}^{op}$ still requires that we consider, in this categories, arrows $r^{op}: c\rightarrow x$, $s^{op}:c\rightarrow y$, $t^{op}:x\rightarrow y$. Importantly, $c$ is still the domain of these arrows (and presumably for these arrows to exist, there must have been arrows $r,s,t$ in $\mathcal{C}$ such that $r:x\rightarrow c$, $s:y\rightarrow c$, and $t:y\rightarrow x$.
In $c/\mathcal{C}^{op}$, a morphism is $t^{op}(r^{op})=s^{op}$, and the objects here are $(r^{op}:c\rightarrow x)$ and $(s^{op}:c\rightarrow y)$. So, when you pass to the opposite cateogry $(c/\mathcal{C}^{op})^{op}$, can you "reach in" to the objects and change their arrow directions (i.e., mutate the objects so that ($r^{op}:c\rightarrow x$) becomes ($r:x\rightarrow c)$)? That's the only way I can get this framework to logically work.
Let's drop the informal terminology of "reaching in", "mutating", "changing arrow directions" and so on and treat this exercise somewhat more formally; this will hopefully clear up the confusion. Namely, let's construct an isomorphism (that is, an invertible functor) $F\colon (c / \mathcal{C}^\mathrm{op})^\mathrm{op} \to \mathcal{C} / c$.
What are the objects of $(c / \mathcal{C}^\mathrm{op})^\mathrm{op}$? They're the same as the objects of $c / \mathcal{C}^\mathrm{op}$, i.e. morphisms of the form $r^\mathrm{op}\colon c \to x$ where $r\colon x \to c$ is a morphism. This gives us an obvious choice for what $F$ should be on objects: we'll define $F(r^\mathrm{op}) = r$, which by definition is an object of $\mathcal{C} / c$.
As for morphisms, what's a morphism in $(c / \mathcal{C}^\mathrm{op})^\mathrm{op}$? Well, it's the ${{-}}^\mathrm{op}$ of a morphism in $(c / \mathcal{C}^\mathrm{op})$, and a morphism in $(c / \mathcal{C}^\mathrm{op})$ from $r^\mathrm{op}\colon c \to x$ to $l^\mathrm{op}\colon c \to y$ is a morphism $h^\mathrm{op}\colon x \to y$ in $\mathcal{C}^\mathrm{op}$ such that $l^\mathrm{op} = h^\mathrm{op} \circ r^\mathrm{op}$, so you could write it as $(h^\mathrm{op})^\mathrm{op}\colon y \to x$. Again it should not come as a surprise that $F$ will be defined on morphisms as $F((h^\mathrm{op})^\mathrm{op}) = h$.
Of course one now needs to check that this actually defines a functor and find an inverse; this I leave as exercise :)