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. $\square$

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

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