For ordinary fibrations is it true that: Given a functor $F \colon C \to D$ with $C$ finitely complete, and a fully faithful functor $U$ such that $F \dashv U$ and F preserving limits. Then F is a fibration?
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Yes, if you either assume also that $F$ is an isofibration or are content with proving that it's a Street fibration. And you only need pullbacks, not a terminal object.
Given $x\in C$ with $F(x)=b$ and a morphism $g:a\to b$ in $D$, let $g^*x$ be the pullback of the unit $\eta_x : x \to U(b) = UF(x)$ along $U(g) : U(a) \to U(b)$. Since $F(\eta_x)$ is an isomorphism (as $U$ is fully faithful) and $F$ preserves pullbacks, $F$ also maps the map $g^*x \to U(a)$ to an isomorphism. And if $F$ is an isofibration, we can modify $g^*x$ to make this map to an equality. Finally, it's straightforward to prove that the other map $g^*x \to x$ is cartesian, using the universal property of the pullback.
answered Jun 2 at 7:48
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$\begingroup$ Did you mean $F$ also maps the map $g^{*}a \to U(a)$ to an isomorphism? So the liftings are not up to equality but isomorphism, so $F$ would be a street fibration? $\endgroup$– SiyaCommented Jun 26 at 17:19
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$\begingroup$ Yes, $g^*x \to U(a)$, I fixed it, thanks. And yes, as I said, if $F$ is not an isofibration, then you can only show that it's a Street fibration. $\endgroup$ Commented Jun 27 at 21:12
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$\begingroup$ One last question, is there a reason why for any morphism $h \colon x \to g^* x$ in $C$, $Fh=h’$? For a morphism $h’ \colon x’ \to a$ in $D$. Or we have isomorphism instead of equality also. $\endgroup$– SiyaCommented Jun 30 at 22:56
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$\begingroup$ Sorry, I don't understand your question. Are you trying to prove that the map $g^*x \to x$ is cartesian? $\endgroup$ Commented Jul 1 at 23:09
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$\begingroup$ Yes, well now I see from the first comment that all the liftings are up to iso. Hence we would have $Fh \cong h'$. Thanks, I will look at the conditions in which $F$ would actually be an isofibration. $\endgroup$– SiyaCommented Jul 2 at 12:27