Slice of opposite category equivalent to coslice of category? Let $\mathcal{C}$ be some category, and $A,B\in\mathcal{C}$.
We have the notions of the slice category $\mathcal{C}/A$ whose objects are morphisms $A'\to A$ and the coslice category $A/\mathcal{C}$ whose objects are morphisms $A\to A'$.
I am pretty sure that $$\mathcal{C}^{\mathrm{op}}/A\equiv A/\mathcal{C}$$ but I'm worried that I might have got an arrow the wrong way around at some point, and that the actual result is $$\mathcal{C}^{\mathrm{op}}/A\equiv (A/\mathcal{C})^{\mathrm{op}}.$$
The first one is what I hope to be true, since I would like it to be the case that a functor $$\mathcal{C}^{\mathrm{op}}/B\to \mathcal{C}^{\mathrm{op}}/A$$ gives a functor $$B/\mathcal{C}\to A/\mathcal{C}$$ and vice versa, but obviously (I hope) if the second statement is the true one then this functor will flip and be $A/\mathcal{C}\to B/\mathcal{C}$.
 A: Slice and coslice categories in a category $C$ both admit a forgetful functor to $C$, which is one way to convince yourself that it's the second one: we have
$$C^{op}/c \cong (c/C)^{op}.$$
These categories both admit a forgetful functor to $C^{op}$.
But your hope is still fine: a functor $F : C^{op}/b \to C^{op}/a$ is the same thing as a functor $F : (b/C)^{op} \to (a/C)^{op}$, which is in turn the same thing as a functor $F^{op} : b/C \to a/C$. Taking opposite categories doesn't flip functors around (it flips natural transformations). 
A: Guess your out of luck indeed it is true that $(A/\mathcal C)^\text{op} \cong \mathcal C^\text{op}/A$
The isomorphism between these two categories is given by 
$$F \colon (A/\mathcal C)^\text{op} \to C^\text{op}/A$$


*

*where 
$$F(f) = f$$ 
for every $f \colon A \to X$ in $\mathcal C$

*and 
$$F(\alpha)=\alpha$$ 
for every $\alpha \colon f \to g$ in $(A/\mathcal C)[f,g]$, with $f \colon A \to X$ and $g \colon A \to Y$.
This functor is clearly well defined on objects because $f \colon A \to X$, that is an object of $A/\mathcal C$, is also an object of $\mathcal C^\text{op}/A$.
On the other hand if $\alpha \in (A/\mathcal C)[f,g]$, with $f \colon A \to X$ and $g \colon A \to Y$ in $\mathcal C$, then $\alpha \colon X \to Y$ and $g=\alpha\circ f$ in $\mathcal C$.
From this, since 
$$g=\alpha\circ f=f \circ^\text{op} \alpha$$ 
it follows that $\alpha \in (\mathcal C^\text{op}/A)[g,f]$ (this shows that the map is contravariant, that is that $F$ is a map from the graph $(A/\mathcal C)^\text{op}$ to $\mathcal C^\text{op}/A$).
Functoriality follows by simple computations and this is an isomorphism since it is bijective both on the objects and the morphisms.
