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$ \newcommand{C}{\mathcal{C} } \newcommand{D}{\mathcal{D} } \newcommand{I}{\mathcal{I} } $ (Full) subcategory $\D$ of the category $\C$ is called reflexive if there exists a localization functor $L : \C \to \D$ and it is left adjoint to the inclusion functor $I : \D \hookrightarrow \C$.

There is theorem:

If $\D$ is a reflective subcategory of $\C$, then $I$ creates all limits that $\C$ admits.

But if this theorem is true, then every diagram $x: \I \to\D$, which admits a limit $y = \lim_{i \in \I} Ix_i $, will also have a limit in $\D$. But as inclusion $I$ is right adjoint, it preserves limits, and so $y = I\Big(\lim_{i \in \I} x_i \Big)$. Then, as $y$ is included into $\C$, it must hold that $\lim_{i \in \I} x_i = Ly$ and $ ILy = y$. But, I don't know how to prove this fact, as $L$ only preserves colimits.

I can prove that $Ly$ indeed forms a cone over $x$ and for any other cone $c$ there exists a morphism $\psi: c \to Ly$, but I don't know how to prove uniqueness.

I construct $\psi$ as $L\phi$, there $\phi: Ic \to y$, which exists by universal property of limit $y$.

What am I missing here? Сan you help me prove uniqueness?

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  • $\begingroup$ Your proof has some problems : in particular, you can't write $\lim_{i \in \I} Ix_i=I\Big(\lim_{i \in \I} x_i \Big)$ if you don't know that $\lim_{i \in \I} x_i$ exists in $\D$ beforehand. And it's a bit problematic to write $y\cong Ly$ since formally $y$ is in $\C$ and $Ly$ is in $\D$. $\endgroup$ – Arnaud D. Aug 6 '18 at 11:41
  • $\begingroup$ It's not a proof, just a line of reasoning in case theorem is already proved. This is an argument of type: "If solution exists, then solution has certain form, hence, I will take this form and prove that it is indeed a solution". By $\lim_{i \in I} x_i= Ly \cong y$ I meant $\lim_{i \in I} x_i = Ly $ and $y = ILy$, sorry if this is too ambiguous. $\endgroup$ – Nik Pronko Aug 6 '18 at 11:55
  • $\begingroup$ Well, I'm really nitpicking, but it can help sometimes ;) I hadn't realised you weren't really asking for a proof, sorry. $\endgroup$ – Arnaud D. Aug 6 '18 at 11:56

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