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edited Mar 5 at 16:12

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If I'm not mistaken, a cosequence of the arguments in Murre's proof is that the inclusion $J^{\text{op}}\colon\mathcal I^{\text{op}}\hookrightarrow \mathcal E ^{\text{op}}$ is a final functor: It's easy to see that if $H\colon \mathcal A\hookrightarrow \mathcal B$ is fully faithful and $\mathcal B$ is filtered, then $H$ is final if and only if for any $b\in\mathcal B$ there exists $a\in\mathcal A$ and a map $b\to Ha$ in $\mathcal B$. In your case this condition is satisfied by point (*) of the proof.

Now use the property of final functors applied to $J^{\text{op}}$: this says (in particular) that if the colimit of $G\colon \mathcal E^{\text{op}}\to \mathcal K$ exists, then also the colimit of $GJ^{\text{op}}\colon \mathcal I^{\text{op}}\to\mathcal K$ exists, and in that case they coincide. In your situation we already know that the colimit indexed on $\mathcal E^{\text{op}}$ exists and is equal to $F$ (since itit is always true, independently of the size of $F$$\mathcal C$, that any functor $F\colon \mathcal C\to \text{Set}$ is the colimit of the diagram indexed on the category of elements), thus you can conclude.

If I'm not mistaken, a cosequence of the arguments in Murre's proof is that the inclusion $J^{\text{op}}\colon\mathcal I^{\text{op}}\hookrightarrow \mathcal E ^{\text{op}}$ is a final functor: It's easy to see that if $H\colon \mathcal A\hookrightarrow \mathcal B$ is fully faithful and $\mathcal B$ is filtered, then $H$ is final if and only if for any $b\in\mathcal B$ there exists $a\in\mathcal A$ and a map $b\to Ha$ in $\mathcal B$. In your case this condition is satisfied by point (*) of the proof.

Now use the property of final functors applied to $J^{\text{op}}$: this says (in particular) that if the colimit of $G\colon \mathcal E^{\text{op}}\to \mathcal K$ exists, then also the colimit of $GJ^{\text{op}}\colon \mathcal I^{\text{op}}\to\mathcal K$ exists, and in that case they coincide. In your situation we already know that the colimit indexed on $\mathcal E^{\text{op}}$ exists (since it is $F$), thus you can conclude.

If I'm not mistaken, a cosequence of the arguments in Murre's proof is that the inclusion $J^{\text{op}}\colon\mathcal I^{\text{op}}\hookrightarrow \mathcal E ^{\text{op}}$ is a final functor: It's easy to see that if $H\colon \mathcal A\hookrightarrow \mathcal B$ is fully faithful and $\mathcal B$ is filtered, then $H$ is final if and only if for any $b\in\mathcal B$ there exists $a\in\mathcal A$ and a map $b\to Ha$ in $\mathcal B$. In your case this condition is satisfied by point (*) of the proof.

Now use the property of final functors applied to $J^{\text{op}}$: this says (in particular) that if the colimit of $G\colon \mathcal E^{\text{op}}\to \mathcal K$ exists, then also the colimit of $GJ^{\text{op}}\colon \mathcal I^{\text{op}}\to\mathcal K$ exists, and in that case they coincide. In your situation we already know that the colimit indexed on $\mathcal E^{\text{op}}$ exists and is equal to $F$ (it is always true, independently of the size of $\mathcal C$, that any functor $F\colon \mathcal C\to \text{Set}$ is the colimit of the diagram indexed on the category of elements), thus you can conclude.

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created Mar 5 at 12:00

499
4
6

If I'm not mistaken, a cosequence of the arguments in Murre's proof is that the inclusion $J^{\text{op}}\colon\mathcal I^{\text{op}}\hookrightarrow \mathcal E ^{\text{op}}$ is a final functor: It's easy to see that if $H\colon \mathcal A\hookrightarrow \mathcal B$ is fully faithful and $\mathcal B$ is filtered, then $H$ is final if and only if for any $b\in\mathcal B$ there exists $a\in\mathcal A$ and a map $b\to Ha$ in $\mathcal B$. In your case this condition is satisfied by point (*) of the proof.

Now use the property of final functors applied to $J^{\text{op}}$: this says (in particular) that if the colimit of $G\colon \mathcal E^{\text{op}}\to \mathcal K$ exists, then also the colimit of $GJ^{\text{op}}\colon \mathcal I^{\text{op}}\to\mathcal K$ exists, and in that case they coincide. In your situation we already know that the colimit indexed on $\mathcal E^{\text{op}}$ exists (since it is $F$), thus you can conclude.