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Given a category $\mathbb{C}$ and its object $C$, the contravariant hom functor based at $C$ is a contravariant functor $hom_{\mathbb{C}}(-,C):\mathbb{C} \to Sets$, where $Sets$ is the category of sets, carrying a morphism $f:A \to B$ in $\mathbb{C}$ to the map $hom_\mathbb{C}(f,C):hom_\mathbb{C}(B,C) \to hom_\mathbb{C}(A,C)$ defined $\alpha \mapsto \alpha{f}$.

As we can use an alternative definition for any contravariant functor using a covariant functor from $\mathbb{C}^{op}$, my notes say as follows.

Given a category $\mathbb{C}$ and its object $C$, the contravariant hom functor based at $C$ is a covariant functor $hom_{\mathbb{C}}(-,C):\mathbb{C}^{op} \to Sets$, where $Sets$ is the category of sets, carrying a morphism $f:B \to A$ in $\mathbb{C}^{op}$ to the map $hom_\mathbb{C}(f,C):hom_\mathbb{C}(A,C) \to hom_\mathbb{C}(B,C)$ defined $\alpha \mapsto \alpha{f}$ (in $\mathbb{C}$).

I strongly believe this is a typo and it should read as follows.

Given a category $\mathbb{C}$ and its object $C$, the contravariant hom functor based at $C$ is a covariant functor $hom_{\mathbb{C}}(-,C):\mathbb{C}^{op} \to Sets$, where $Sets$ is the category of sets, carrying a morphism $f:B \to A$ in $\mathbb{C}^{op}$ to the map $hom_\mathbb{C}(f,C):hom_\mathbb{C}(B,C) \to hom_\mathbb{C}(A,C)$ defined $\alpha \mapsto \alpha{f}$ (in $\mathbb{C}$).

Any help will be greatly appreciated.

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    $\begingroup$ You are completely right. $\endgroup$ – asdq Aug 14 '18 at 11:30
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    $\begingroup$ You are right, if the map $f: B\to A$ is in $\mathbb{C}^{op}$, then it's $hom(B,C)\to hom(A,C)$ $\endgroup$ – Max Aug 14 '18 at 11:30
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    $\begingroup$ @asdq Thank you so much $\endgroup$ – James Aug 14 '18 at 11:45
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    $\begingroup$ @Max Thank you for the confirmation $\endgroup$ – James Aug 14 '18 at 11:45
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    $\begingroup$ @Roland write it out to see : it can only be one of them (and you have to specify whether you're talking about $\mathbb{C}$ or $ \mathbb{C}^{op}$ $\endgroup$ – Max Oct 29 '18 at 9:58

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