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Earlier today, I was thinking: "Oh, an $R$-module is just an additive functor $R \rightarrow \mathbf{Ab}.$" Anyway, I had a bit of a read over at nLab, and it says:

For any small $\mathbf{Ab}$-enriched category $R$, the enriched presheaf category $[R^{op},\mathbf{Ab}]$ is, of course, $\mathbf{Ab}$-enriched. If $R$ is a ring, as above, then $[R^{op},\mathbf{Ab}]$ is the category of $R$-modules.

This is gonna sound dumb, but I don't get why its $R^{op}$ rather than $R$. Explanation, anyone?

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  • $\begingroup$ It probably has to do with the order of multiplication/composition in the ring and whether the ring is acting on the module on the left or the right. Earlier in the nLab page you link to, it says "For instance, we can talk about (left and right) modules over a ringoid $R$, which can be defined as $\text{Ab}$-enriched functors $R\to\text{Ab}$ and $R^{\text{op}}\to\text{Ab}$." $\endgroup$ – Alex Kruckman Nov 23 '14 at 6:29
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$[R^{op}, \text{Ab}]$ is the category of right $R$-modules, which is equivalently the category of left $R^{op}$-modules. The reason to prefer taking right modules here is the same reason why presheaves are contravariant functors and not covariant functors: it's so that the Yoneda embedding, which in this case is $R \to [R^{op}, \text{Ab}]$, is covariant. In particular, the unit right $R$-module, namely $R$ itself equipped with right multiplication, has endomorphism $R$ this way. (The unit left $R$-module, by contrast, has endomorphism ring $R^{op}$.)

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  • $\begingroup$ According to the Principle of conservation of the number of ops, the number of ops remains constants no matter how hard one tries to come up with cunning definitions to avoid them... In fact, the harder one tries, the more annoying the ops become. $\endgroup$ – Mariano Suárez-Álvarez Nov 23 '14 at 9:52

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