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The usual decategorification $K(C)$ of a monoidal category $C$ is by taking the isomorphic classes of $C$ as objects. The resulting decategorification $K(C)$ is a monoid.

There is another decategorification that takes the morphisms as objects instead. The result is also a monoid. Is there a name for and literature on this?

EDIT: Let me clarify after Boris' comment. Start with a pointed monoidal category $C$. Then the set of arrows $Arr(C)$ is a monoid with the tensor product and the identity arrow.

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  • $\begingroup$ Sorry, do you consider the morphisms as objects and the commutative triangles as morphisms (and obtain a new category)? Either do you consider the morphisms as elements with the operation of composition (and obtain a semigroup rather than monoid)? $\endgroup$ Nov 4, 2011 at 14:08
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    $\begingroup$ This is the 'decategorification' of the monoidal category of diagrams with shape $\circ\rightarrow\bullet$ $\endgroup$ Nov 4, 2011 at 14:35
  • $\begingroup$ I'm not sure if Fernando has answered what you want, but I felt like I should add that $Arr(C)$ has another monoidal structure given by the pushout product...$f_i:X_i\rightarrow Y_i$ has $f_0\square f_1: X_0\otimes Y_1 \coprod_{X_0\otimes X_1} Y_0\otimes X_1 \rightarrow Y_0\otimes Y_1$. I wonder what you get when you take isomorphism classes with respect to this rather than the standard $\otimes$ in $Arr(C)$ $\endgroup$ Nov 4, 2011 at 16:31
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    $\begingroup$ Another decategorification is $C \mapsto \mathrm{End}(1_C)$. $\endgroup$ Nov 4, 2011 at 17:17
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    $\begingroup$ Yes, Fernando answered my original question. The decategorification in my edit is the functor category $[\cdot\to\cdot,C]$. Thank you Boris for allowing me to clarify my intended decategorification. As pointed other, there are really several (different) decategorifications. $\endgroup$
    – user2529
    Nov 5, 2011 at 5:56

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