Let consider a complete regular category $\mathscr C$. For $x:X\to A$ and $y:Y\to A$ we write $x\equiv y$ if there exists regular epimorphisms $u:W\to X$ and $v:W\to Y$ such that $x\circ u=y\circ v$.

Let $\Phi$ be a diagram in $\scr C$ indexed by a small category $\scr I$. Assume that for each object $n$ of $\scr I$ is given a morphism $\phi_n:D_n\to\Phi_n$ such that for each $i:n\to m$ in $\scr I$ we have $\phi_m\equiv\Phi(i)\circ \phi_n$.

Let $\lambda_n:\lim\Phi\to\Phi_n$ be a limit cone.

Question: there exists a morphism $\phi:D\to\lim\Phi$ such that $\phi_n\equiv\lambda_n\circ\phi$?

Clearly the answer is yes if we require $\phi_m=\Phi(i)\circ\phi_n$.

  • $\begingroup$ What is $D$? It appears without any explanation, first with a subscript and then without? Perhaps the first $D_n$ should be without a subscript, so that you are essentially defining a cone to $\Phi$ where equality is replace by this weaker relation? $\endgroup$ – Nex Oct 20 '16 at 19:25
  • $\begingroup$ $D$ is the domain of $\phi$. It's no related to $D_n$. In other words, each $\phi_n$ and $\phi$ can have different domains; thus $\phi_n$ not necessarily defines a cone to $\Phi$. $\endgroup$ – Fabio Lucchini Oct 21 '16 at 6:10

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