Show that $C/c \cong (c/(C^{op}))^{op}$ I have trouble seeing that $C/c \cong (c/(C^{op}))^{op}$. This is what I have managed so far:

What am I doing wrong here? It seems to me that to get to $(c/(C^{op}))^{op}$, would have to reverse the $h$, and then it no longer looks like $C/c$

(Edit)
This is just to sum up what my mistake was, in case others fall in the same trap: What I missed was the fact, that (see diagram) in $c/C^{op}$, not only $f$ and $g$ get reversed (ie. $f \rightarrow f^{op}$ etc), but $h$ too becomes $h^{op}$. Why did I not see that? Well, my thinking was along the lines "$C = \cdot \xrightarrow{f} \cdot$", so "$C^{op} = \cdot \xrightarrow{f^{op}} \cdot$", and to contruct the slice category, you create morphisms, let's call them $h$ ..., - but of course, those morphisms are between objects in the categories that are getting sliced (if that's the right term), so the same relationship applies here: $h \rightarrow h^{op}$.
Thanks for kindly helping me see this!
 A: The objects of $(c/\mathcal{C}^{\mathrm{op}})^{\mathrm{op}}$ are exactly the objects of $c/\mathcal{C}^{\mathrm{op}}$, which are morphisms $c \to x$ in $\mathcal{C}^{\mathrm{op}}$, which are morphisms $x \to c$ in $\mathcal{C}$ (marked with 'op', if that's your convention). So an object of $(c/\mathcal{C}^{\mathrm{op}})^{\mathrm{op}}$ is $f^{\mathrm{op}}$ for some $f \in \mathcal{C}/c$.
Given $f : x \to c$ and $g : y \to c$ in $\mathcal{C}$, a morphism $f^{\mathrm{op}} \to g^{\mathrm{op}}$ in $(c/\mathcal{C}^{\mathrm{op}})^{\mathrm{op}}$ is $h^{\mathrm{op}}$ for some morphism $h : g^{\mathrm{op}} \to f^{\mathrm{op}}$ in $c/\mathcal{C}^{\mathrm{op}}$, which is a morphism $h : y \to x$ in $\mathcal{C}^{\mathrm{op}}$ such that $h \circ g^{\mathrm{op}} = f^{\mathrm{op}}$, which in turn is $k^{\mathrm{op}}$ for some $k : x \to y$ in $\mathcal{C}$ such that $g \circ k = f$. In summary, a morphism $h : f^{\mathrm{op}} \to g^{\mathrm{op}}$ in $(c/\mathcal{C}^{\mathrm{op}})^{\mathrm{op}}$ is $(k^{\mathrm{op}})^{\mathrm{op}}$ for some $k : f \to g$ in $\mathcal{C}/c$.
Thus you have a map $\mathcal{C}/c \to (c/\mathcal{C}^{\mathrm{op}})^{\mathrm{op}}$ defined on objects by $f \mapsto f^{\mathrm{op}}$ and on morphisms by $h \mapsto (h^{\mathrm{op}})^{\mathrm{op}}$.
You now need to check that these assignments describe a functor and that it is an isomorphism.
