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First, we note that the coherator on an affine scheme is given by the associated quasi-coherent sheaf to the global section (cf. Thomason-Trobaugh, Appendix B. 14).

Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$ and $f = \mathrm{id}_S$. Then, $f_{\#}$ is exactly the coherator $Q_S:\mathsf{Mod}(\mathcal{O}_S)\to \mathsf{QCoh}(S)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$.

Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$. Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$.

Here, we give a simple counterexample: Let $(R, \mathfrak{m})$ be a DVR. Write

• $S:\overset{\mathrm{def}}{=} \mathrm{Spec}(R)$, $K$ for the field of fractions of $R$, $U :\overset{\mathrm{def}}{=} \mathrm{Spec}(K)$, $j:U\hookrightarrow S$ for the natural open immersion,
• $X :\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}}\mathrm{Spec}(R/\mathfrak{m}^i)$, $F:\overset{\mathrm{def}}{=} \mathcal{O}_X$, and
• $f:X\to S$ for the natural morphism.

Then, $f_{\#}F$ is equal to the associated quasi-coherent sheaf to $\prod_{i\in \mathbb{N}}R/\mathfrak{m}^i$. Hence, $j^*f_{\#}F\neq 0$. On the other hand, since $X\times_{\mathrm{Spec}(K)} S = \emptyset$, the natural morphism $0\neq j^*f_{\#}F \to f_{\#}'{j'}^*F = 0$ is not an isomorphism.

On a related note, we can prove the following claim:

Claim. Let $f:X\to S$ be a morphism of schemes and $F$ a quasi-coherent sheaf on $X$. Assume that for any open immersion $j:U \hookrightarrow S$, the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is an isomorphism.

Then, $f_*F$ is a quasi-coherent sheaf on $S$.

Proof. Write $Q_{(-)}(-)$ for the coherator. Let $j:U\hookrightarrow S$ be an open immersion. Since $f'_{\#}(-) = Q_U(f'_*(-))$, it follows from the assumption that \begin{align} f_{\#}F(U) &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, j^*f_{\#}F) \xrightarrow{\sim} \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, f'_{\#}{j'}^*F) \\ &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U,Q_U(f_*'{j'}^*F)) = \mathrm{Hom}_{\mathsf{Mod}(\mathcal{O}_U)}(\mathcal{O}_U,f_*'{j'}^*F) \\ &= \Gamma(U, f_*'{j'}^*F) = F(f^{-1}(U)) = f_*F(U). \end{align} This implies that the natural morphism $f_{\#}F \to f_*F$ is an isomorphism of sheaves. Thus, in particular, $f_*F$ is quasi-coherent. ◻︎

By the above claim, we conclude that if $f_*F$ is not quasi-coherent, then there exists an open immersion $j:U\hookrightarrow S$ such that the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is not an isomorphism.

First, we note that the coherator on an affine scheme is given by the associated quasi-coherent sheaf to the global section (cf. Thomason-Trobaugh, Appendix B. 14).

Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$. Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$.

Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$.

Here, we give a simple counterexample: Let $(R, \mathfrak{m})$ be a DVR. Write

• $S:\overset{\mathrm{def}}{=} \mathrm{Spec}(R)$, $K$ for the field of fractions of $R$, $U :\overset{\mathrm{def}}{=} \mathrm{Spec}(K)$, $j:U\hookrightarrow S$ for the natural open immersion,
• $X :\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}}\mathrm{Spec}(R/\mathfrak{m}^i)$, $F:\overset{\mathrm{def}}{=} \mathcal{O}_X$, and
• $f:X\to S$ for the natural morphism.

Then, $f_{\#}F$ is equal to the associated quasi-coherent sheaf to $\prod_{i\in \mathbb{N}}R/\mathfrak{m}^i$. Hence, $j^*f_{\#}F\neq 0$. On the other hand, since $X\times_{\mathrm{Spec}(K)} S = \emptyset$, the natural morphism $0\neq j^*f_{\#}F \to f_{\#}'{j'}^*F = 0$ is not an isomorphism.

On a related note, we can prove the following claim:

Claim. Let $f:X\to S$ be a morphism of schemes and $F$ a quasi-coherent sheaf on $X$. Assume that for any open immersion $j:U \hookrightarrow S$, the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is an isomorphism.

Then, $f_*F$ is a quasi-coherent sheaf on $S$.

Proof. Write $Q_{(-)}(-)$ for the coherator. Let $j:U\hookrightarrow S$ be an open immersion. Since $f'_{\#}(-) = Q_U(f'_*(-))$, it follows from the assumption that \begin{align} f_{\#}F(U) &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, j^*f_{\#}F) \xrightarrow{\sim} \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, f'_{\#}{j'}^*F) \\ &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U,Q_U(f_*'{j'}^*F)) = \mathrm{Hom}_{\mathsf{Mod}(\mathcal{O}_U)}(\mathcal{O}_U,f_*'{j'}^*F) \\ &= \Gamma(U, f_*'{j'}^*F) = F(f^{-1}(U)) = f_*F(U). \end{align} This implies that the natural morphism $f_{\#}F \to f_*F$ is an isomorphism of sheaves. Thus, in particular, $f_*F$ is quasi-coherent. ◻︎

By the above claim, we conclude that if $f_*F$ is not quasi-coherent, then there exists an open immersion $j:U\hookrightarrow S$ such that the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is not an isomorphism.

First, we note that the coherator on an affine scheme is given by the associated quasi-coherent sheaf to the global section (cf. Thomason-Trobaugh, Appendix B. 14).

Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$ and $f = \mathrm{id}_S$. Then, $f_{\#}$ is exactly the coherator $Q_S:\mathsf{Mod}(\mathcal{O}_S)\to \mathsf{QCoh}(S)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$.

Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$. Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$.

Here, we give a simple counterexample: Let $(R, \mathfrak{m})$ be a DVR. Write

• $S:\overset{\mathrm{def}}{=} \mathrm{Spec}(R)$, $K$ for the field of fractions of $R$, $U :\overset{\mathrm{def}}{=} \mathrm{Spec}(K)$, $j:U\hookrightarrow S$ for the natural open immersion,
• $X :\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}}\mathrm{Spec}(R/\mathfrak{m}^i)$, $F=\mathcal{O}_X$, and
• $f:X\to S$ for the natural morphism.

Then, $f_{\#}F$ is equal to the associated quasi-coherent sheaf to $\prod_{i\in \mathbb{N}}R/\mathfrak{m}^i$. Hence, $j^*f_{\#}F\neq 0$. On the other hand, since $X\times_{\mathrm{Spec}(K)} S = \emptyset$, the natural morphism $0\neq j^*f_{\#}F \to f_{\#}'{j'}^*F = 0$ is not an isomorphism.

We can prove the following.

Claim. Let $f:X\to S$ be a morphism of schemes and $F$ a quasi-coherent sheaf on $X$. Assume that for any open immersion $j:U \hookrightarrow S$, the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is an isomorphism.

Then, $f_*F$ is a quasi-coherent sheaf on $S$.

Proof. Write $Q_{(-)}(-)$ for the coherator. Let $j:U\hookrightarrow S$ be an open immersion. Since $f'_{\#}(-) = Q_U(f'_*(-))$, it follows from the assumption that \begin{align} f_{\#}F(U) &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, j^*f_{\#}F) \xrightarrow{\sim} \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, f'_{\#}{j'}^*F) \\ &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U,Q_U(f_*'{j'}^*F)) = \mathrm{Hom}_{\mathsf{Mod}(\mathcal{O}_U)}(\mathcal{O}_U,f_*'{j'}^*F) \\ &= \Gamma(U, f_*'{j'}^*F) = F(f^{-1}(U)) = f_*F(U). \end{align} This implies that the natural morphism $f_{\#}F \to f_*F$ is an isomorphism of sheaves. In particular, $f_*F$ is quasi-coherent. ◻︎

First, we note that the coherator on an affine scheme is given by the associated quasi-coherent sheaf to the global section (cf. Thomason-Trobaugh, Appendix B. 14).

Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$ and $f = \mathrm{id}_S$. Then, $f_{\#}$ is exactly the coherator $Q_S:\mathsf{Mod}(\mathcal{O}_S)\to \mathsf{QCoh}(S)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$.

Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$. Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$.

Here, we give a simple counterexample: Let $(R, \mathfrak{m})$ be a DVR. Write

• $S:\overset{\mathrm{def}}{=} \mathrm{Spec}(R)$, $K$ for the field of fractions of $R$, $U :\overset{\mathrm{def}}{=} \mathrm{Spec}(K)$, $j:U\hookrightarrow S$ for the natural open immersion,
• $X :\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}}\mathrm{Spec}(R/\mathfrak{m}^i)$, $F:\overset{\mathrm{def}}{=} \mathcal{O}_X$, and
• $f:X\to S$ for the natural morphism.

Then, $f_{\#}F$ is equal to the associated quasi-coherent sheaf to $\prod_{i\in \mathbb{N}}R/\mathfrak{m}^i$. Hence, $j^*f_{\#}F\neq 0$. On the other hand, since $X\times_{\mathrm{Spec}(K)} S = \emptyset$, the natural morphism $0\neq j^*f_{\#}F \to f_{\#}'{j'}^*F = 0$ is not an isomorphism.

On a related note, we can prove the following claim:

Claim. Let $f:X\to S$ be a morphism of schemes and $F$ a quasi-coherent sheaf on $X$. Assume that for any open immersion $j:U \hookrightarrow S$, the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is an isomorphism.

Then, $f_*F$ is a quasi-coherent sheaf on $S$.

Proof. Write $Q_{(-)}(-)$ for the coherator. Let $j:U\hookrightarrow S$ be an open immersion. Since $f'_{\#}(-) = Q_U(f'_*(-))$, it follows from the assumption that \begin{align} f_{\#}F(U) &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, j^*f_{\#}F) \xrightarrow{\sim} \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, f'_{\#}{j'}^*F) \\ &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U,Q_U(f_*'{j'}^*F)) = \mathrm{Hom}_{\mathsf{Mod}(\mathcal{O}_U)}(\mathcal{O}_U,f_*'{j'}^*F) \\ &= \Gamma(U, f_*'{j'}^*F) = F(f^{-1}(U)) = f_*F(U). \end{align} This implies that the natural morphism $f_{\#}F \to f_*F$ is an isomorphism of sheaves. Thus, in particular, $f_*F$ is quasi-coherent. ◻︎

By the above claim, we conclude that if $f_*F$ is not quasi-coherent, then there exists an open immersion $j:U\hookrightarrow S$ such that the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is not an isomorphism.

Even though $f$ is qcqs and $S$ is qcqs, the natural morphism $j^*f_{\#}\to f_{\#}'{j'}^*$ is not an isomorphism. Here is an example:

Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$ and $f = \mathrm{id}_S$. Then, $f_{\#}$ is exactly the coherator $Q_S:\mathsf{Mod}(\mathcal{O}_S)\to \mathsf{QCoh}(S)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$.

Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$. Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$ (cf. Thomason-Trobaugh, Appendix B. 14).

By using this observation, we can conclude the following:

Lemma. Let $F$ be an $\mathcal{O}_S$-module such that for any $f\in A$, if we write $S_f :\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$, then the natural morphism $Q_S(F)|_{S_f}\to Q_{S_f}(F|_{S_f})$ is an isomorphism.

Then, $F$ is a quasi-coherent sheaf.

Since $Q_S(F)$ is quasi-coherent, it holds that $\Gamma(S_f, Q_S(F)|_{S_f}) = \Gamma(S, Q_S(F))_f$. Since $Q_S(F) = \Gamma(S, F)\tilde{ \ }$, it holds that $\Gamma(S, Q_S(F)) = \Gamma(S, F)$. Hence, it holds that $\Gamma(S_f, Q_S(F)|_{S_f}) = \Gamma(S, F)_f$. Thus, it follows from the assumption that the natural morphism $\Gamma(S,F)_f\to \Gamma(S_f, F)$ is an isomorphism.

Since the family of open subschemes $\left\{ S_f \,\middle|\, f\in A\right\}$ is an open base of $S$, this implies that for any $x\in S$, the morphism $Q_S(F)_x\to F_x$ induced by taking stalks of the natural morphism $Q_S(F)\to F$ is an isomorphism. Hence, in particular, $Q_S(F)\to F$ is an isomorphism of sheaves.

This implies that $F$ is a quasi-coherent sheaf. ◻︎

By the above Lemma, we can conclude that for any non-quasi-coherent $\mathcal{O}_S$-module $F$, there exists an element $f\in A$ such that the natural morphism $Q_S(F)|_{S_f}\to Q_{S_f}(F|_{S_f})$ is not an isomorphism. In particular, if we write $j: S_f \hookrightarrow S$ for the natural open immersion, then the natural morphism $j^*\mathrm{id}_{S,\#}F \to \mathrm{id}_{S_f, \#}{j'}^*F$ is not an isomorphism.

A more concrete counterexample:

Let $A$ be a DVR. Write $K$ for the field of fractions of $A$. Define an $\mathcal{O}_S$-module $F$ as follows: $F(\mathrm{Spec}(A)) = 0, F(\mathrm{Spec}(K)) = K$.

Then, $Q_{\mathrm{Spec}(A)}(F) = 0$, $Q_{\mathrm{Spec}(K)}(F|_{\mathrm{Spec}(K)}) = \tilde{K}$, and, moreover, the natural morphism $0\to K$ is not an isomorphism.

Note that if $g:T\to S$ is a morphism of schemes, then the scheme-theoretic image of $g$ is equal to the closed subscheme corresponding to the quasi-coherent ideal sheaf $Q_S(\mathrm{ker}(g^{\#}:\mathcal{O}_S\to g_*\mathcal{O}_T))\subset \mathcal{O}_S$. The above $F$ is a special case of this: $T:\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}} \mathrm{Spec}(A/(\pi^i))$, where $\pi\in A$ is a uniformizer, and $g: T \to \mathrm{Spec}(A)$ for the canonical morphism of schemes. Then, $F = \mathrm{ker}(\mathcal{O}_{\mathrm{Spec}(A)}\to g_*\mathcal{O}_T)$, and the scheme theoretic image of $g$ is equal to $\mathrm{Spec}(A)$.

First, we note that the coherator on an affine scheme is given by the associated quasi-coherent sheaf to the global section (cf. Thomason-Trobaugh, Appendix B. 14).

Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$ and $f = \mathrm{id}_S$. Then, $f_{\#}$ is exactly the coherator $Q_S:\mathsf{Mod}(\mathcal{O}_S)\to \mathsf{QCoh}(S)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$.

Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$. Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$.

Here, we give a simple counterexample: Let $(R, \mathfrak{m})$ be a DVR. Write

• $S:\overset{\mathrm{def}}{=} \mathrm{Spec}(R)$, $K$ for the field of fractions of $R$, $U :\overset{\mathrm{def}}{=} \mathrm{Spec}(K)$, $j:U\hookrightarrow S$ for the natural open immersion,
• $X :\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}}\mathrm{Spec}(R/\mathfrak{m}^i)$, $F=\mathcal{O}_X$, and
• $f:X\to S$ for the natural morphism.

Then, $f_{\#}F$ is equal to the associated quasi-coherent sheaf to $\prod_{i\in \mathbb{N}}R/\mathfrak{m}^i$. Hence, $j^*f_{\#}F\neq 0$. On the other hand, since $X\times_{\mathrm{Spec}(K)} S = \emptyset$, the natural morphism $0\neq j^*f_{\#}F \to f_{\#}'{j'}^*F = 0$ is not an isomorphism.

We can prove the following.

Claim. Let $f:X\to S$ be a morphism of schemes and $F$ a quasi-coherent sheaf on $X$. Assume that for any open immersion $j:U \hookrightarrow S$, the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is an isomorphism.

Then, $f_*F$ is a quasi-coherent sheaf on $S$.

Proof. Write $Q_{(-)}(-)$ for the coherator. Let $j:U\hookrightarrow S$ be an open immersion. Since $f'_{\#}(-) = Q_U(f'_*(-))$, it follows from the assumption that \begin{align} f_{\#}F(U) &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, j^*f_{\#}F) \xrightarrow{\sim} \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, f'_{\#}{j'}^*F) \\ &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U,Q_U(f_*'{j'}^*F)) = \mathrm{Hom}_{\mathsf{Mod}(\mathcal{O}_U)}(\mathcal{O}_U,f_*'{j'}^*F) \\ &= \Gamma(U, f_*'{j'}^*F) = F(f^{-1}(U)) = f_*F(U). \end{align} This implies that the natural morphism $f_{\#}F \to f_*F$ is an isomorphism of sheaves. In particular, $f_*F$ is quasi-coherent. ◻︎