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$\textbf{Definition}$

A category $\mathscr{C}$ is complete if every (covariant) functor $F:\mathscr{D}\rightarrow \mathscr{C}$ has a limit with $\mathscr{D}$ a small category.

Let $\mathscr{C}$ be a complete category. Then, is it true that $\mathscr{C}$ is a preordered set if $\mathscr{C}$ is small? (So that the term "complete category" is only meaningful to large categories.)

Assume that $\mathscr{C}$ is small. Now take $\mathscr{D}$ as the discrete category whose objects are arrows in $\mathscr{C}$. Take two objects $C_1,C_2$ in $\mathscr{C}$ and suppose there are two distinct morphisms $f,g:C_1\rightarrow C_2$. Let $\Delta_{C_2}:\mathscr{D}\rightarrow \mathscr{C}$ be the (covariant) constant functor. Since $\mathscr{C}$ is small and complete, there exists the limit $L$ of this constant functor. However, since there are two distinct morphisms $f,g:C_1\rightarrow C_2$, we can construct $2^{|\mathscr{D}|}$ distinct cones, hence $2^{|\mathscr{C}|}$ distinct morphisms from $C_1$ to $L$. Since there are only $|\mathscr{D}|$ morphisms in $\mathscr{C}$, this is a contradiction. Consequently, if $\mathscr{C}$ is small, then it is merely a preordered set.

Is my argument correct?

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Yes, this is correct and is a well-known argument. Any small complete category is a preorder, and so usually you only talk about complete categories in the context of large categories.

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  • $\begingroup$ I wondered if I misunderstood something. Thank you! $\endgroup$ – Rubertos Jan 26 '17 at 23:43

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