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Given $f,g$ as morphisms of objects in $C$, we denote them as $f',g'$ for $C^{op}$, and $$f\circ_C g = g'\circ_{C^{op}} f'.$$ For a contravariant functor $F$ from $C$ to $D$, we have $f:x\rightarrow y, g: y\rightarrow z$ $$F(g\circ_C f) = F(f) \circ_D F(g)$$ Why is this the same as a functor $F$ from $C^{op}$ to $D$? So given $$F(g' \circ_{C^{op}} f') = F(g')\circ_D F(f')$$ why is $$F(f) \circ_D F(g)= F(g')\circ_D F(f')?$$

Edit: If what I wrote doesn't make sense, can you point me to a reference? I see people say "the opposite category is used to define contravariant in a uniform way", but there is not any detail.

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    This what I found from here http://math.stackexchange.com/questions/438694/definition-of-opposite-category in the answer. – 2017-01-23
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    Yeah ok, dumb comment of me. But your notation is confusing. Let $f\in \text{Hom}_{C^{\text{op}}}(X,Y)$, by definition $f\in \text{Hom}_{C}(Y,X)$, no need to use $f'$. – 2017-01-23
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    **Please** make sure that you avoid typos **specially** in titles, which is the first thing people see of your question. We do judge books by their covers, you know... – 2017-01-23

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Using your notation, note that $$F(g' \circ_{C^{\mathrm{op}}} f') = F(f \circ_C g)$$ and not $F(g \circ_C f)$, and so you're trying to compare the wrong $D$-morphisms on the 8th line of your question (i.e. the equation after 'why is').

Treating $F$ as a covariant functor $C^{\mathrm{op}} \to D$, we have $$F(g \circ_C f) = F(f' \circ_{C^{\mathrm{op}}} g') = F(f') \circ_D F(g')$$ which is the same result as when we considered $F$ as a contravariant function $C \to D$, since $f=f'$ and $g=g'$.

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    Ah I see, that typo got me really confused, but I did not know that $f = f'$ and $g=g'$, I will read more about this. Thank you. – 2017-01-23
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    Morphisms in $C^{\mathrm{op}}$ are morphisms in $C$, but their associated domain and codomain are the other way round. Specifically, suppose $f$ is a morphism in $C$ with domain $A$ and codomain $B$. Then $f$ is also a morphism in $C^{\mathrm{op}}$, but this time its domain is $B$ and its codomain is $A$. That's why I say $f=f'$. – 2017-01-23