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Jan 6, 2015 at 18:00 comment added Tim Campion Even if $\zeta$ is associative, the composition in $\zeta * C$ is typically not associative, so it is not a category. Suppose that $f$ is an isomorphism in $C$ which is not an identity, and let $g$ be an arrow in $C$ with the same codomain. Then consider the two ways to compose $(f,i)\circ (f^{-1},j) \circ (g,k)$. If you do the left composition first, you get $(g,k)$. If you do the right composition first, you get $(g,ijk)$ where I'm writing the operation of $\zeta$ as concatenation. Typically $ijk \neq k$ so these do not agree.
Apr 7, 2013 at 20:11 history edited Salvo Tringali CC BY-SA 3.0
Changed the title, rephrased some parts
Apr 7, 2013 at 9:48 comment added Salvo Tringali Thanks, Theo, for your comment. Yes, sorry, I forgot to say that $\zeta$ must be associative, in the categorial case. Let me edit and fix it.
Apr 7, 2013 at 9:22 history edited Salvo Tringali CC BY-SA 3.0
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Apr 7, 2013 at 3:05 comment added Theo Johnson-Freyd I am likely confused about some aspect of your construction, but I don't see how $\zeta \ast C$ can be a category unless $\zeta: \alpha\times\alpha \to \alpha$ is associative. If $\zeta$ is associative, then this is (almost, but not quite) the operation of "base change" of a category: if $C$ is any category, and $(\alpha,\zeta,a\in\alpha)$ any associative algebra, then "$C \otimes \alpha$" is the category with the same objects as $C$ and morphisms $\hom_{C\otimes \alpha}(x,y) = \hom_C(x,y)\times \alpha$. I have no useful comments about colimits.
Apr 6, 2013 at 23:35 history edited Salvo Tringali CC BY-SA 3.0
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Apr 6, 2013 at 20:53 history edited Salvo Tringali CC BY-SA 3.0
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Apr 6, 2013 at 20:45 history asked Salvo Tringali CC BY-SA 3.0