Given an equational theory $\mathbb{T}$, it is known that the category fp-$\mathbb{T}$-mod of all finitely-presented models of $\mathbb{T}$ in $Sets$ has finite colimits. But is it known whether this category also has finite limits?
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$\begingroup$ @Vladimir: nope. Try writing it down. $\endgroup$– Qiaochu YuanCommented Oct 14, 2016 at 3:20
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$\begingroup$ @Qiochu right, of course, otherwise my question math.stackexchange.com/questions/1930595 about kernel pairs of finitely presented models being finitely generated would have a very easy answer... $\endgroup$– Vladimir SotirovCommented Oct 14, 2016 at 3:41
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1 Answer
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No, not in general. For instance, take groups. The free group $F_2$ on two generators is finitely presented, as is the free abelian group $\mathbb{Z}^2$ on two generators. But the equalizer in groups of the canonical homomorphism $F_2\to \mathbb{Z}^2$ and the trivial homomorphism is the subgroup $[F_2,F_2]$, which is free on infinitely many generators. If an equalizer of these maps existed in finitely presented groups, there would be a single finitely presented group $G$ with a map $\varphi:G\to [F_2,F_2]$ such that every map from a finitely presented group to $[F_2,F_2]$ factors uniquely through $\varphi$. But this would imply $\varphi$ is surjective, since every element of $[F_2,F_2]$ can be in the image of a map from a finitely presented group (namely $\mathbb{Z}$). Since $[F_2,F_2]$ is not finitely generated, no such $\varphi$ can exist.
answered Oct 14, 2016 at 3:26