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If $C$ is a category, we want to construct a new category $C^{op}$. To do so, we need to describe its objects, the set of morphisms between each of its objects and its composition rule $\circ_{C^{op}}$.
The objects of $C^{op}$ are just the objects of $C$.
If $a$ and $b$ are two objects of $C^{op}$, then the set $\hom_{C^{op}}(a,b)$ is defined to be $\hom_C(b,a)$. Notice that this makes sense, for $a$ and $b$ are objects of $C$, so it makes sense to talk about $\hom_C(b,a)$.
If $a$, $b$ and $c$ are three objects of $C^{op}$ and $f\in\hom_{C^{op}}(a,b)$ and $g\in\hom_{C^{op}}(b,c)$, then the above definition means that $f\in\hom_C(b,a)$ and $g\in\hom_C(c,b)$, so it makes sense to compute the composition $f\circ_C g$ in $C$, which gives an element of $\hom_C(c,a)$. This last set is, by definition $\hom_{C^{op}}(a,c)$. We define the composition of $C^{op}$ so that $$g\circ_{C^{op}}f:= f\circ_Cg.$$
Let us give a concrete example. Suppose $C$ is the category which has exactly two objects, which are called $a$ and $b$ and three morphisms in all: the two identities and one morphism $f:a\to b$. Then $C^{op}$ has again $\{a,b\}$ as set of objects, and has exactly three morphisms: the two identities and one morphism from $b$ to $a$, which is $f$.
A slightly less silly example is the following: suppose $C$ is a category which has exactly one object, which I will call $x$. Then the set $\hom_C(x,x)$ is a set endowed with an associative multiplication admitting an identity element: it is what one calls a monoid. The opposite category $C^{op}$ has, again, exactly one object, which is still $x$, and has a monoid $\hom_{C^{op}}(x,x)$. Can you describe this monoid in terms of $\hom_C(x,x)$?