I'm working on Exercise 1.2.i. in this book and I don't have any idea how to start.
Drawing out the diagrams on page 8 and flipping them around a couple times I've vaguely come to the conclusion that these categories are "the same", but the details are hazy and I'm not sure what to make of it. Here's my closest attempt to formally showing the isomorphism.
Let $C$ be a category with object $c$. We want to show that $C/c \cong (c/(C^\text{op}))^\text{op}$. Since every object $f: x \to c$ in $C/c$ and in $(c/C^\text{op})^\text{op}$ is a morphism $f: x \to c$ in $C$, and every morphism $\alpha: f \to g$ (where $g: y \to c \in C$) is a morphism $\alpha: x \to y$ in $C$, it seems like you could define a functor $F: C/c \to (c/C^\text{op})^\text{op}$ where $Ff = f$ and $F\alpha = \alpha$, which is certainly an isomorphism. I feel like I need to prove something here, but I'm not sure what or how. Is this functor valid & does it define the isomorphism we're asked for?
I found a similar question that had been asked a while ago where someone gave a very thorough answer which I think involves more information than mine but I'm not sure. Someone also commented another question I had about this problem - just based on the way it's mapping objects and morphisms, this functor seems a lot like the identity, but obviously its source and target are not the same. Is there a name for functors like these, or just different types of isomorphisms in general? Some feel special, like the identity and whatever isomorphism there is between $C$ and $C^\text{op}$.
$\newcommand{\op}{{^\mathsf{op}}}\newcommand{\ob}{\mathsf{Ob}}$I'm not sure that you've been super careful here. An object $f$ of $C/c$ "is" an arrow $f:x\to c$ in $C$. There is an associated arrow $f\op:c\to x$ in the category $C\op$, so there is this object $f\op$ in $c/C\op$. So on objects we have a well-defined assignment $f\mapsto f\op$ that runs $\ob(c/C)\to\ob(c/C\op)=\ob((c/C\op)\op)$.
But you want to make this work on arrows too. Say $\alpha\in(C/c)(f,g)$; that is, $\alpha:x\to y$ is an arrow in $C$ and $g:y\to c$ has $g\circ\alpha=f$. How does this translate to $c/C\op$? Well, in $C\op$ we have the equation $\alpha\op\circ g\op=f\op:c\to x$. In other words, $\alpha\op$ is an arrow $g\op\to f\op$ in $c/C\op$. That's going in the wrong direction and that's the thing I'm not convinced you've thought about.
But, $(\alpha\op)\op$ (using two different senses of "$\op$", ugh, but the notation is limited) is then an arrow $f\op\to g\op$ in $(c/C\op)\op$. This is the correct direction now; I invite you to continue the exercise by checking the functor $C/c\to(c/C\op)\op$, $f\mapsto f\op$ and $\alpha\mapsto(\alpha\op)\op$ is actually a functor (preserves identities and composition) and is an isomorphism of categories.