Why is the "opposite category" operation a functor? I have seen some question in this site saying that the "opposite category" opration is a functor. But from the definition of a functor, we require $F(g\circ f)=F(g)\circ F(f)$. What I feel confused about is that for $f: A\to B, g: B\to C,\;$ $F(f)$ is from $B$ to $A$ and $F(g)$ is from $C$ to $B$ so we cannot compose them as $F(g)\circ F(f)$. So why does this operation satisfy the definition of a functor? Am I misunderstanding something?
Thanks a lot!
 A: I'm afraid I don't really understand what you're trying to say is a problem; I think you are somehow confused about what it means to say that the opposite functor operation is a functor.  Let me just walk through how that works.
To avoid some confusion, let me fix some notation.  If $\mathcal{C}$ is a category, we write $\mathcal{C}^{op}$ for its opposite category.  If $X$ is an object of $\mathcal{C}$, we write $X^o$ for $X$ considered as an object of $\mathcal{C}^{op}$.  If $f:X\to Y$ is a morphism of $\mathcal{C}$, we write $f^o:Y^o\to X^o$ for the corresponding morphism of $\mathcal{C}^{op}$.
The claim now is that there is a functor $T:\mathtt{Cat}\to\mathtt{Cat}$ which on objects is given by $T(\mathcal{C})=\mathcal{C}^{op}$ for each category $\mathcal{C}$.  To verify this, we need to specify what $T$ does on morphisms: that is, if $F:\mathcal{C}\to\mathcal{D}$ is a functor, we need to define a functor $T(F):\mathcal{C}^{op}\to\mathcal{D}^{op}$.  The definition is simple: on objects we define $T(F)(X^o)=F(X)^o$ and on morphisms we define $T(F)(f^o)=F(f)^o$.
Let's check that this $T(F)$ really is a functor. If $f^o:Y^o\to X^o$ is a morphism of $\mathcal{C}^{op}$, then $f:X\to Y$ in $\mathcal{C}$, so $F(f):F(X)\to F(Y)$ and $F(f)^o:F(Y)^o\to F(X)^o$.  That is, $T(F)(f^o)=F(f)^o$ is indeed a morphism from $T(F)(Y^o)=F(Y)^o$ to $T(F)(X^o)=F(X)^o$.
We can also check that $T(F)$ preserves identities and composition.  For identities, we have $$T(F)(1_{X^o})=T(F)(1_X^o)=F(1_X)^o=1_{F(X)}^o=1_{F(X)^o}.$$ For composition, suppose $f^o:Y^o\to X^o$ and $g^o:Z^o\to Y^o$ are two composable morphisms in $\mathcal{C}^{op}$.  Then $$T(F)(f^og^o)=T(F)((gf)^o)=F(gf)^o=(F(g)F(f))^o=F(f)^oF(g)^o=T(F)(f^o)T(F)(g^o).$$  Here we use the fact that composition in the opposite category is defined by $f^og^o=(gf)^o$.
So, for any functor $F:\mathcal{C}\to\mathcal{D}$, we have defined a functor $T(F):\mathcal{C}^{op}\to\mathcal{D}^{op}$.  The only thing that remains to be checked is that this operation $T$ preserves identities and composition of functors: $T(1_{\mathcal{C}})=1_{\mathcal{C}^{op}}$ and $T(FG)=T(F)T(G)$ if $F:\mathcal{C}\to\mathcal{D}$ and $G:\mathcal{B}\to\mathcal{C}$ are functors.  This is tedious but straightforward to check and does not appear to be your point of confusion so I will omit the details.
