Show that $\text{C}/c \cong (c/(\text{C}^{\text{op}}))^{\text{op}}$ I'm self studying from Category Theory in Context by Emily Riehl and have encountered the following question regarding slice categories:

Let $\text{C}$ be a category and $c$ be an object in $\text{C}$. Show that $\text{C}/c \cong (c/(\text{C}^{\text{op}}))^{\text{op}}$ 

The exercise appears in an chapter that introduces the notion of opposite categories and duality proofs. 

My intuition of why the statement is true comes from the following observation:
Let $x,y$ be objects in $\mathbf{C}$ (or equivalently $\mathbf{C}^{\text{op}}$). Let $g^{\text{op}} : c \to y$ and $f^{\text{op}} : c \to x$ be objects in $c/(\mathbf{C}^{\text{op}})$. By the definition of slice category, there exists a morphism $h^{\text{op}} : y \to x$ in $\mathbf{C}^{\text{op}}$ such that the diagram below commutes:
$\hskip2in$ 
Now the 'corresponding diagram' in the opposite category $(c/(\mathbf{C}^{\text{op}}))^{\text{op}}$ would be:
$\hskip2in$  
which, as I understand it, would also be commutative. Now, to me, this indicates:


*

*that the objects in $(c/(\mathbf{C}^{\text{op}}))^{\text{op}}$ are the same as the objects in $\mathbf{C}/c$, namely, a morphism in $\mathbf{C}$ from an object $z \in \mathbf{C}$ to $c$.

*the commutativity of the seconds diagram describes morphisms $h$ in both $\mathbf{C}/c$ and $(c/(\mathbf{C}^{\text{op}}))^{\text{op}}$

*if you reverse the arrows again you return to the same diagram. This makes me think you could construct a map between morphisms by first making a map from $ \mathbf{C}/c \to (\mathbf{C}/c)^{\text{op}} ( \cong c/(\mathbf{C}^{\text{op}}) )$ by $h \mapsto h^{op}$



Now, what I am really looking for is a way to structure a proof. I feel like I have some intuition but am very unsure of how to write it down in a concise way. I am new to category theory so I'm still very much learning how to construct these types of proofs using duality; I feel like what I have written is maybe overkill but I'm trying to be as explicit as possible to make sure I understand everything.
In particular, I don't know how I would get a concrete correspondence between morphisms in the two categories that respects associativity of composition etc.
Any help or pointers would be appreciated!
 A: As often with these kinds of questions, you need to start with the definitions. You want to prove that two categories are isomorphic. What does that mean? It might sound stupid, but that's the first step: it means you need to find an isomorphism between the two.
What's an isomorphism of categories? It's a functor $F : \mathbf{C}/c \to (c/\mathbf{C}^{op})^{op}$ which admits an inverse functor, i.e. a functor $G : (c/\mathbf{C}^{op})^{op} \to \mathbf{C}/c$ such that $F \circ G$ and $G \circ F$ are both identities. So now you know what to do: you need to find these two functors and prove they are inverse to each other.
So first you want to find $F : \mathbf{C}/c \to (c/\mathbf{C}^{op})^{op}$. Now is the right time to use your intuition that you developed in your question.


*

*A functor should first be defined on objects. So suppose you have an object of $\mathbf{C}/c$, say $f : x \to c$. You want to build an object of $(c/\mathbf{C}^{op})^{op}$ out of this. There aren't tons of ways of doing this. Basically, the only option you have is to consider $F(f) := f^{op} : c \to x$. You already saw this in your question.

*Now you want to define $F$ on morphisms. Morphisms are the commutative diagrams that you wrote in your question: given $(f : x \to c)$, $(g : y \to c)$ objects of $\mathbf{C}/c$, a morphism is some $h : x \to y$ such that $g \circ h = f$. You want to define $F(h)$. You know it has to be a morphism in $(c/\mathbf{C}^{op})^{op}$ from $g^{op} : c \to y$ to $f^{op} : c \to x$. You saw this in your question too: you can just define $F(h) := h^{op}$.


The next step is to prove that this is indeed a functor. So you need to check that $F$ maps identities to identities, and distributes over composition. For example, what is the identity of the object $f : x \to c$ in $\mathbf{C}/c$? It's simply $\operatorname{id}_x : x \to x$. You need to check if $F(\operatorname{id}_x)$ is the identity of $f^{op} : c \to x$ in $c/\mathbf{C}$. And it's true, because $F(\operatorname{id}_x) = \operatorname{id}_x^{op}$ is the identity of $x$ in $\mathbf{C}^{op}$. You need to do the same thing for composition, I'll let you do it.
The next step is to find the inverse functor $G : (c/\mathbf{C}^{op})^{op}$. Think about it for a moment. You will see that the definition is very similar to $F$; on objects you send $f^{op} : c \to x$ to $G(f^{op}) := f : x \to c$, and on morphisms you do the same thing, $G(h^{op}) := h$.
You then need to prove that $G$ is a functor, which is again a simple exercise in definitions. Finally, you want to prove that $F$ and $G$ are inverse to each other, i.e. $F \circ G = \operatorname{id}$ and $G \circ F = \operatorname{id}$. But that's pretty obvious: $G(F(f)) = G(f^{op}) = f$ (on objects), and similarly in the other direction. So now you're done.
