The Stacks project

Hijyen belgesi, işletmelerin hijyen standartlarına uygun çalıştığını belgeleyen önemli bir sertifikadır. Bu belge, özellikle gıda sektöründe faaliyet gösteren işletmelerin güvenilirliğini artırırken, müşteri memnuniyetini de sağlar. Hijyen sertifikası ile işletmenizin yasal düzenlemelere uygun olduğunu kanıtlayabilir, sağlıklı bir ortamda hizmet sunduğunuzu gösterebilirsiniz. hijyen belgesi

In the same line to Comment 9051 above, I am a bit confused with the replacement written there. Maybe something like the following is better? (Note the added $H^i$ on the left hand sides, and the $[-i]$ on the right hand sides.)

"Here is another way to think of the cup product of $\xi$ and $\eta$. Namely, we can write $$\begin{equation} H^i(R\Gamma(X ,K)) = \operatorname{Hom}_{D(\mathcal{O}_X)}(\mathcal{O}_X[-i], K) \end{equation}$$

and $$\begin{equation} H^j(R\Gamma(X ,M)) = \operatorname{Hom}_{D(\mathcal{O}_X)}(\mathcal{O}_X[-j], M) \end{equation}$$

because $\operatorname{Hom}(\mathcal{O}_X, -) = \Gamma(X, -)$"

Actually, one could add another a sixth point into the list of equivalences:

1. There is an isomorphism $\beta:X\oplus Z\cong Y$ such that $(1_X,\beta,1_Z):(X,X\oplus Z,Z,(1,0),(0,1),0)\to(X,Y,Z,f,h,g)$ is a triangle isomorphism.

Implication 6$\Rightarrow$1 is trivial. Implication (1-5)$\Rightarrow$6 is what is the current proof does to show that for any right inverse $s:Z\to Y$ of $g$ the map $f\oplus s:X\oplus Z\to Y$ is an isomorphism.

I think this result could be expanded/rephrased in this way:

Lemma. Let $\mathcal{D}$ be a pre-triangulated category.

• For any objects $X$, $Z$ of $\mathcal{D}$ the triangle $(X,X\oplus Z,Z,(1,0),(0,1),0)$ is distinguished.

• Let $(X,Y,Z,f,g,h)$ be a distinguished triangle in $\mathcal{D}$. The following are equivalent:

  1. $h=0$,
  2. $f$ is a monomorphism,
  3. $f$ has a left inverse,
  4. $g$ is an epimorphism, and
  5. $g$ has a right inverse.

Moreover, if these equivalent conditions hold, then for any right inverse $s:Z\to Y$ of $g$ the map $f\oplus s:X\oplus Z\to Y$ is an isomorphism.

Proof. The proof of the last assertion is written in the already existing proof. The first point follows from TR1 and Lemma 13.4.10. We prove the second point. Let $W$ be an object of $\mathcal{D}$. The sequences $$\begin{gather*} \operatorname{Hom}(W,Z)\xrightarrow{h_*} \operatorname{Hom}(W,X[1])\xrightarrow{f[1]_*} \operatorname{Hom}(W,Y[1])\\ \operatorname{Hom}(X[1],W)\xrightarrow{h^*} \operatorname{Hom}(Z,W)\xrightarrow{g^*} \operatorname{Hom}(Y,W) \end{gather*}$$ are exact. By Yoneda, $h=0$ if and only $h_*=0$ for all $W$ and/or $h^*=0$ for all $W$. That is, $h=0$ if and only if $f[1]_*$ is injective for all $W$ and/or $g^*$ is injective for all $W$. In other words, $h=0$ if and only if $f[1]$ is monic and/or $g$ is epic. Since $f$ is monic iff $f[1]$ is, this shows 1$\Leftrightarrow$2$\Leftrightarrow$4. It is clear that 3$\Rightarrow$2 and $5\Rightarrow$4. Now assume 1. Then plugging $(X,Y,Z,f,g,h)$ into $\operatorname{Hom}(-,X)$ and $\operatorname{Hom}(Z,-)$ gives exactness of $$\begin{gather*} \operatorname{Hom}(Y,X)\xrightarrow{f^*} \operatorname{Hom}(X,X)\xrightarrow{h^*=0} 0\\ \operatorname{Hom}(Z,Y)\xrightarrow{g_*} \operatorname{Hom}(Z,Z)\xrightarrow{h_*=0} 0 \end{gather*}$$ Thus there are lifts of $1_X\in\operatorname{Hom}(X,X)$ and $1_Z\in\operatorname{Hom}(Z,Z)$ along these maps, i.e., a left inverse for $f$ and a right inverse for $g$, respectively. This finishes the proof of the second point. $\square$

Is there maybe an unstated fact that if $X \to S$ is pseudocoherent then $E \in D(\mathcal{O}_X)$ is pseudocoherent iff it's pseudocoherent relative to $S$? Maybe this is in the Stacks Project somewhere and I missed it. As it stands I don't think you've explained why (B) is equivalent to $S$-perfect when $X \to S$ is finitely presented and flat because in the definition of $S$-perfect you say $E$ is pseudocoherent and in (B) you say $E$ is pseudocoherent relative to $S$.

The x in the addendum is not the same as the x in the main statement, Better to say morphism y -> z in A (for every y,z in Ob(A) Furthermore, if such an object 0 exists, then a morphism α: y→ z factors through 0 if and only if α=0.

$B_{univ}$ may also be divided by $1 - \sum b^le_l$, and there are typos about "$\mathfrak{a}$ vs. $\mathfrak{a}'$".

Here's the proof: The result follows from Categories, Lemma 4.17.2 applied to the observation that the functor $s^*:Y/S\to X/S$, sending an object $Y\to Y'$ to an object $X\xrightarrow{s}Y\to Y'$, is cofinal. To show that $s^*$ is cofinal, we check the conditions of #9830. Suppose $X\to X'$ is an object of $X/S$. Then there are morphisms $Y\to Y'\leftarrow X'$ such that the square commutes and $X\to Y''$ is in $S$. Since $S$ is saturated, also $Y\to Y'$ lies in $S$. On the other hand, suppose we have a commutative diagram $$\begin{matrix} X&\xrightarrow{s}&Y\\ \downarrow&&\downarrow\\ X'&\rightrightarrows&Y' \end{matrix}$$ Since $X/S$ is filtered, there is a morphism $Y'\to Y''$ coequalizing $X'\rightrightarrows Y'$ and such that the composite $X\to X'\rightrightarrows Y'\to Y''$ lies in $S$. Since $S$ is saturated, $Y'\to Y''$ lies also in $S$, thus $Y\to Y'\to Y''$ is an object in $Y/S$ and we win.

There is another result in the spirit of Lemma 4.19.3 that is occasionally useful. It goes likes this:

[KS, Proposition 3.2.2]. Let $\mathcal{J}$ be a filtered category. Let $F:\mathcal{J}\to\mathcal{I}$ be a functor. Then $F$ is cofinal if and only if the following two conditions hold:

1. For each $i\in \mathcal{I}$ there exists $j\in\mathcal{J}$ and a morphism $s:i\to F(j)$.

2. For any $i\in\mathcal{I}$, any $j\in\mathcal{J}$, and any pair of parallel morphisms $s,s':i\rightrightarrows F(j)$ in $\mathcal{I}$, there exists a morphism $t:j\to k$ in $\mathcal{J}$ such that $F(t)\circ s= F(t)\circ s'$.

Moreover, if $F$ is cofinal, then $\mathcal{I}$ is filtered.

I believe there is a small reduction step missing after step 6. Namely, in order to get $f(X)$ with coefficients in $R$, one needs $\mathfrak{n}$ to be maximal. But this is not automatically the case (for example, let $R$ be a complete DVR with fraction field $K$ and let $x\in K$ be any element outside of $R$; then $S=K$ and its maximal ideal does not lie over the maximal ideal of $R$).

To fix this, one must replace $R$ by $R_\mathfrak{n}$, $S$ by $R_\mathfrak{n}[x]\subset K$, and $\mathfrak{m}$ by $\mathfrak{m}S_\mathfrak{m}\cap R_\mathfrak{n}[x]$. Since $S\subset R_\mathfrak{n}[x]\subset S_\mathfrak{m}$, this is indeed a maximal ideal of $R_\mathfrak{n}[x]$ lying over $\mathfrak{n}R_\mathfrak{n}$ and with localization $S_\mathfrak{m}$.

Sorry, I meant $A^\wedge\cap K$ and $A^\wedge\otimes_A K$, respectively.

The proof seems to miss a logical step. Namely, in order to show that $B$ is a domain one needs that $a$ doesn't have a root in the fraction field $K$ of $A$ (or am I overlooking an obvious argument?). For this one needs to use that $A^\hat\cap K=A$ inside $A^\hat\otimes_A K$ and that $A^\hat\otimes_A K$ is reduced (which is clear since $A$ is analytically unramified).

The proof seems to miss a logical step. Namely, in order to show that $B$ is a domain one needs that $a$ doesn't have a root in the fraction field $K$ of $A$ (or am I overlooking an obvious argument?). For this one needs to use that $A^\hat\cap K=A$ inside $A^\hat\otimes_A K$ and that $A^\hat\otimes_A K$ is reduced (which is clear since $A$ is analytically unramified).

Typo: in the statement of the lemma, it should read $g(U) \subset U'$.

Typoe: in (4) of Lemma 0A0K, the $I$'s should be $\mathcal{I}$ I believe.

Typo in the second line of the lemma. The assumption should read: If $N$ is flat, then ...

We can actually generalize this result to:

Lemma. Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system in $\mathcal{D}$ compatible with the triangulated structure. Let $\mathcal{D}'\subset\mathcal{D}$ be a full pre-triangulated subcategory. Suppose at least one of the following assertions hold:

1. For every morphism $X'\to X$ in $S$ with $X'\in\mathcal{D}'$ there is a morphism $X\to X''$ in $S$ with $X''\in\mathcal{D}'$.

2. For every $X\in\mathcal{D}$ there is $X\to X'$ in $S$ with $X'\in\mathcal{D}'$.

Then $S'=S\cap\operatorname{Mor}(\mathcal{D}')$ is a multiplicative system in $\mathcal{D}'$ compatible with the triangulated structure, and if $S$ is saturated then so is $S'$. Moreover, the triangulated functor $(S')^{-1}\mathcal{D}'\to S^{-1}\mathcal{D}$ coming from Lemma 13.5.7 is fully faithful. Furthermore, if 2 holds, then $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ is an equivalence of pre-triangulated categories.

Proof. By #9820, we must only show MS5 and MS6 for $S'$. But this follows from the same axioms for $S$ and the fact that $\mathcal{D}'$ is a full and triangulated sub-category. $\square$

Hello! I found your article very informative and read it with great interest. I discovered this blog only recently and I am glad I did. I will definitely read more articles. Congratulations on the blog and regards!

@#9817 Okay, sorry for the several comments. Here's the most general statement:

Lemma. Let $\mathcal{C}$ be a category. Let $S$ be a left multiplicative system in $\mathcal{C}$. Let $\mathcal{C}'\subset\mathcal{C}$ be a full subcategory. Suppose any of the following assertions hold: 1. For every morphism $X'\to X$ in $S$ with $X'\in\mathcal{C}'$ there is a morphism $X\to X''$ in $S$ with $X''\in\mathcal{C}'$. 2. For every $X\in\mathcal{C}$ there is $X\to X'$ in $S$ with $X'\in\mathcal{C}'$.

Then $S'=S\cap\operatorname{Mor}(C')$ is a left multiplicative system in $\mathcal{C}'$ and the functor $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ coming from Lemma 4.27.8 is fully faithful. Moreover, if 2 holds, then $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ is an equivalence.

Clearly 2 implies 1, and the proof I wrote in #9817 can be done only by assuming only 1. Essential surjectivity of $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ by assuming 2 is just the fact that $X\to X'\in S$ becomes an iso in $S^{-1}\mathcal{C}$.