The Stacks project

It took me a little long to parse this proof so I will spell out the details for everyone's convenience: Suppose $\psi_1,\psi_2:x\rightrightarrows x'$ are morphisms in $\mathcal{X}_U$ such that $\rho=F_U\psi_1=F_U\psi_2:F_Ux\to F_Ux'$ in $\mathcal{Y}_U$. Write $y=Fx$ and let $\phi:\operatorname{id}_U^*y\to Fx$ be the induced strongly cartesian morphism (it lives in $\mathcal{Y}_U$). The object $y\in\mathcal{Y}_U$ induces a morphism $\mathcal{C}/U\to\mathcal{Y}$ of fibered categories over $\mathcal{C}$. Write $\phi'=\rho\circ\phi$. Then $\psi_1,\psi_2:(x,\phi)\rightrightarrows(x',\phi')$ are parallel morphisms in the fiber category of $(\mathcal{C}/U)\times_\mathcal{Y}\mathcal{X}$ over the object $\operatorname{id}_U$ of $\mathcal{C}/U$ (Lemma 4.42.1). But this fiber category is a setoid (Lemma 4.40.2), hence $\psi_1=\psi_2$.

Sorry, I just realized $\mathcal{Y}$ was required to be fibred in groupoids, thus $\phi$ is always an iso.

I think in (1) we want “$\phi:f^*y\to F(x)$ is an isomorphism in $\mathcal{Y}_V$,” to match the given definition in Lemma 4.32.3 (and also make sense of $\phi^{-1}$ in (2)).

@#5627: I want to add an observation/caveat maybe relevant to this discussion: If $\mathcal{B}\to\mathcal{C}$ is a Grothendieck fibration (Definition 4.33.5) and $\mathcal{A}\simeq\mathcal{B}$ is an equivalence of categories, then the composite $\mathcal{A}\simeq\mathcal{B}\to\mathcal{C}$ is in general not a Grothendieck fibration, but a Street fibration. In particular, a category $\mathcal{A}$ equivalent to the $2$-fibre product of Lemma 4.33.10 might not be fibered over $\mathcal{C}$.

Nonetheless, the day is saved if the equivalence is in the $2$-category of categories over $\mathcal{C}$, Lemma 4.33.8.

OK, I added an internal reference, see this.

OK, thanks, Fixed here.

Here's a generalization:

[Vis2, Proposition 3.28]. Let $p_\mathcal{G}:\mathcal{G}\to\mathcal{C}$ be a fibered category, $\mathcal{F}$ another category, $F:\mathcal{F}\to\mathcal{G}$ a functor. Then $\mathcal{F}$ is fibered in sets (resp., in groupoids) over $\mathcal{G}$ if and only if it is fibered in sets (resp., in groupoids) over $\mathcal{C}$ via the composite $p_\mathcal{G}\circ F:\mathcal{F}\to\mathcal{G}$.

Proof. Firstly, we have #11409. On the other hand, as the proof of [Vis2, Proposition 3.28] says, “one sees that the fiber of $\mathcal{F}$ over an object $U$ of $\mathcal{C}$ is the disjoint union, as a category, of the fibers of $\mathcal{F}$ over all the objects of $\mathcal{G}$ over $U$; these fiber are groupoids, or sets, if and only if their disjoint union is.” $\square$

Here's a generalization:

[Vis2, Proposition 3.28]. Let $p_\mathcal{G}:\mathcal{G}\to\mathcal{C}$ be a fibered category, $\mathcal{F}$ another category, $F:\mathcal{F}\to\mathcal{G}$ a functor. Then $\mathcal{F}$ is fibered over $\mathcal{G}$ if and only if it is fibered over $\mathcal{C}$ via the composite $p_\mathcal{G}\circ F:\mathcal{F}\to\mathcal{G}$.

Proof. One direction is Lemma 4.33.12. For the other direction, use that every arrow in $\mathcal{G}$ is strongly cartesian (see Lemma 4.35.2 and Definition 4.35.1) and #11349. $\square$

Is there a place where I can find the ommited details from proof of lemma 9.6.9?

Vibe coding for interactive learning is transforming the way students, educators, and professionals engage with digital education. By combining creativity, collaboration, and hands-on coding experiences, Vibe coding for interactive learning creates an environment where users can actively participate instead of passively consuming information. This modern learning approach helps individuals build practical skills while maintaining motivation and long-term engagement. Visit our Website:https://mexty.ai/

I think the proof could be the following:

Suppose the $2$-fibre product of fibered categories $\mathcal{X}\times_\mathcal{S}\mathcal{Y}$ has the description given in Lemma 4.32.3, and suppose the $2$-fibre product of (ordinary) categories $\mathcal{X}_U\times_{\mathcal{S}_U}\mathcal{Y}_U$ has the description given in Example 4.31.3 (see Lemma 4.31.4). Then we have a functor $(\mathcal{X}\times_\mathcal{S}\mathcal{Y})_U\to\mathcal{X}_U\times_{\mathcal{S}_U}\mathcal{Y}_U$ that on objects acts as $(U,x,y,f)\mapsto(x,y,f)$ and that on morphisms acts as $((a,b):(U,x,y,f)\to(U,x',y',f'))\longmapsto((a,b):(x,y,f)\to(x',y',f'))$. By construction, this functor is an isomorphism of categories. $\square$

Dear Professor,

I am reading Lemma 7.6 in the section on differential graded algebra, and I think there may be two small typographical issues in the proof.

In the last line of the proof, it says

$$b\circ b'=0$$

as a map of modules. However, in the statement of the lemma the maps are

$$b:L_1\to L_2,\qquad b':L_2\to L_3.$$

So the composition that is defined, and also the one needed for the conclusion, should be

$$b'\circ b=0.$$

Also, just before that, the proof writes something like

$$\operatorname{Ker}((b')^n)\supset \operatorname{Im}(K_2\to L_2).$$

It seems to me that the superscript (n) should either be removed, giving

$$\operatorname{Ker}(b')\supset \operatorname{Im}(K_2\to L_2),$$

or the right-hand side should also be written degreewise as

$$\operatorname{Ker}((b')^n)\supset \operatorname{Im}(K_2^n\to L_2^n)$$

for every $n$.

Thus I think the end of the proof should read essentially:

$$\operatorname{Im}(b)\subset \operatorname{Im}(K_2\to L_2)\subset \operatorname{Ker}(b'),$$

and hence

$$b'\circ b=0.$$

Please let me know if I have misunderstood the convention here.

Best regards, Zhou JiaWei

Dear Professor,

I am reading Lemma 7.6 in the section on differential graded algebra, and I think there may be two small typographical issues in the proof.

In the last line of the proof, it says

[ b\circ b'=0 ]

as a map of modules. However, in the statement of the lemma the maps are

[ b:L_1\to L_2,\qquad b':L_2\to L_3. ]

So the composition that is defined, and also the one needed for the conclusion, should be

[ b'\circ b=0. ]

Also, just before that, the proof writes something like

[ \operatorname{Ker}((b')^n)\supset \operatorname{Im}(K_2\to L_2). ]

It seems to me that the superscript (n) should either be removed, giving

[ \operatorname{Ker}(b')\supset \operatorname{Im}(K_2\to L_2), ]

or the right-hand side should also be written degreewise as

[ \operatorname{Ker}((b')^n)\supset \operatorname{Im}(K_2^n\to L_2^n) ]

for every (n).

Thus I think the end of the proof should read essentially:

[ \operatorname{Im}(b)\subset \operatorname{Im}(K_2\to L_2)\subset \operatorname{Ker}(b'), ]

and hence

[ b'\circ b=0. ]

Please let me know if I have misunderstood the convention here.

Best regards, Zhou JiaWei

In tag 07VF, near the end we can read "Observe that the claim is clear for $n≤c$ as $f=f′$ mod $I^{c+1} N$". Could it be that what is meant is "Observe that the claim is clear for $n≤c$ as $g=g′$ mod $I^{c+1} N$"? since this part of the proof only uses $g$ and $g'$, and the claim would follow from the statment I suggest.

In tag 07VF, near the end we can read "Observe that the claim is clear for $n≤c$ as $f=f′modIc+1N$". Could it be that what is meant is "Observe that the claim is clear for $n≤c$ as $g=g′modIc+1N$"? since this part of the proof only uses $g$ and $g'$, and the claim would follow from the statment I suggest.

Yes, of course. I could argue that the result that universally closed morphisms are quasi-compact is shown only later and so cannot be used in this spot. We also in general don't formulate lemmas which are purely logical consequences of a combination of other lemmas and we still use those lemmas (by just citing the lemmas involved and letting the reader do the logic required).

Yes, you are right. Going to leave as is.

Thanks and fixed here.

Fixed here.

Fixed here.