Three spans of a set of objects

1. (1) Write \(\mathcal L=\langle\mathcal S\rangle^{\mathrm{GR}}_{\widehat{\mathcal D}}\). For the compact objects \(s[n]\) (\(s\in\mathcal S\), \(n\in\mathbb Z\)), Neeman's lemma says the objects compact in \(\mathcal L\) form their thick span [Neeman 1992, Lemma 2.2]. Coproducts in \(\mathcal L\) are computed in \(\widehat{\mathcal D}\), so these are exactly the objects of \(\mathcal D\cap\mathcal L\).
2. (2) Let \(\iota:\mathcal L\hookrightarrow\widehat{\mathcal D}\). The left orthogonal \({}^{\perp}(\mathcal S^{\perp})\) contains \(\mathcal S\) and is closed under colimits, so it contains \(\mathcal L\). Conversely, \(\mathcal L\) is the smallest full subcategory containing every \(s[n]\) and closed under extensions and small colimits (that subcategory is closed under \([-1]\), hence stable), so it is presentable [HA, Prop. 1.4.4.11(2)] and \(\iota\) has a right adjoint \(R\) [HTT, Cor. 5.5.2.9]. The counit of \(X\in\widehat{\mathcal D}\) has cofiber \(C\in\mathcal S^{\perp}\): $$\iota RX\longrightarrow X\longrightarrow C,\qquad\Map(s,\iota RX)\xrightarrow{\ \sim\ }\Map(s,X)\quad(s\in\mathcal S).$$ If \(X\in{}^{\perp}(\mathcal S^{\perp})\), then \(\Map(X,C)=0=\Map(\iota RX,C)\), so \(\Map(C,C)=0\). Hence \(C=0\) and \(X\simeq\iota RX\in\mathcal L\).
3. (3) \(\mathcal S^{\perp_{\mathcal D}}=\mathcal S^{\perp_{\widehat{\mathcal D}}}\cap\mathcal D\) has fewer test objects, so its left orthogonal in \(\mathcal D\) is larger.

1. Let \(\mathcal A\) be such a category, \(X\in\mathcal A\), and \(I\) a set larger than the set of morphisms of the homotopy category \(h\mathcal A\). Coproducts in \(\mathcal A\) are coproducts in \(h\mathcal A\): $$\operatorname{Hom}_{h\mathcal A}\Big(\textstyle\bigoplus_I X,\,X\Big)\cong\operatorname{End}_{h\mathcal A}(X)^I.$$
2. If \(\operatorname{End}(X)\) had two elements, the right side would have at least \(2^{|I|}\gt|I|\) elements, more than \(h\mathcal A\) has morphisms. So \(\operatorname{End}(X)=\{0\}\), \(\mathrm{id}_X=0\) and \(X\simeq0\). This is the dual of Freyd's argument that a small complete category is a preorder [Freyd, Ch. 3, Exercise D].

1. Choose an open disc \(B\ni a\) with \(B\cap\Sigma_M=\varnothing\), and let \(p:B\to\mathrm{pt}\). The counit is an isomorphism, so \(M|_B\) is constant: $$p^*p_*(M|_B)\xrightarrow{\ \sim\ }M|_B,\qquad M|_B\simeq p^*M_a.$$ Both sides are exact in \(M\), so by induction on the number of nonzero \(\mathcal H^nM\) it suffices to treat one local system \(L\). On the contractible \(B\), \(L\) is constant and \(H^{\gt0}(B,L)=0\), so \(p_*L=L_a\) and the counit is an isomorphism.
2. \(i_a^!\) depends only on \(M|_B\), and \(M_a\simeq\bigoplus_nH^n(M_a)[-n]\) is a finite sum, so it suffices to compute \(i_a^!\underline k_B\). It is local cohomology at \(a\); since \(B\simeq\mathrm{pt}\), \(B\setminus a\simeq S^1\) and restriction is an isomorphism on \(H^0\): $$i_a^!\underline k_B=\operatorname{fib}\big(R\Gamma(B,k)\to R\Gamma(B\setminus a,k)\big)=\operatorname{fib}\big(k\to k\oplus k[-1]\big)=k[-2].$$

Proof of the Proposition

1. Probes are costalks. \(i_{a*}\) is left adjoint to \(i_a^!\), so $$\Map_{\mathcal D}(\delta_a,M)\simeq\Map_{D(k)}(k,i_a^!M)\simeq i_a^!M.$$
2. Finite supports split. Let \(\Sigma\subset\mathbb C\) be finite with \(M_x=0\) for \(x\notin\Sigma\). Then \(M\) vanishes on \(\mathbb C\setminus\Sigma\), and the localization triangle and discreteness of \(\Sigma\) give $$M\simeq\bigoplus_{b\in\Sigma}i_{b*}M_b,\qquad i_a^!\,i_{b*}V\simeq\begin{cases}V,&a=b,\\0,&a\ne b.\end{cases}$$
3. Part (a). Let \(M\in\mathcal S^{\perp_{\mathcal D}}\), so \(i_a^!M=0\) for every \(a\) by step 1. gives \(M_a=0\) for \(a\notin\Sigma_M\). Step 2 with \(\Sigma=\Sigma_M\) then gives \(M_a\simeq i_a^!M=0\) for \(a\in\Sigma_M\). So \(M=0\), and $$\langle\mathcal S\rangle^{\mathrm M}_{\mathcal D}={}^{\perp_{\mathcal D}}0=\mathcal D.$$
4. Supports are thick. The stalk functor \(M\mapsto M_x=i_x^*M\) is exact and carries retracts to retracts. So for a cofiber sequence \(A\to B\to C\) and a retract \(M\) of \(N\): $$\operatorname{supp}M[1]=\operatorname{supp}M,\qquad\operatorname{supp}C\subseteq\operatorname{supp}A\cup\operatorname{supp}B,\qquad\operatorname{supp}M\subseteq\operatorname{supp}N.$$ The objects of finite support therefore form a thick subcategory containing \(\mathcal S\): $$\operatorname{Thick}_{\mathcal D}(\mathcal S)\subseteq\{M\in\mathcal D:\operatorname{supp}M\ \text{finite}\}.$$
5. Part (b). Conversely, if \(\operatorname{supp}M\) is finite, step 2 writes \(M\) as a finite sum of objects \(i_{b*}M_b\simeq\bigoplus_n\delta_b^{\oplus\dim H^n(M_b)}[-n]\), and a finite sum is the cofiber of a zero map. Finally \(\underline k_x=k\ne0\) for every \(x\): $$\operatorname{supp}\underline k_{\mathbb A^1}=\mathbb C\ \text{ is infinite},\qquad\underline k_{\mathbb A^1}\notin\operatorname{Thick}_{\mathcal D}(\mathcal S).$$

1. Objects of \(D^b_c(\mathbb A^1)\) are compact in \(\widehat{\mathcal D}\), so \(\Map(\delta_a,-)\) and \(\Map(\underline k_{\mathbb A^1},-)\) commute with the filtered colimit.
2. The sets \(T\ni a\) are cofinal, and for them \(j_T^*\delta_a=0\): $$\Map(\delta_a,j_{T*}\underline k_{U_T})\simeq\Map(j_T^*\delta_a,\underline k_{U_T})=0.$$
3. \(U_T\) is the plane minus \(|T|\) points, and restriction is an isomorphism on \(H^0\): $$\Map(\underline k_{\mathbb A^1},j_{T*}\underline k_{U_T})\simeq R\Gamma(U_T,k)\simeq k\oplus k^{|T|}[-1].$$

The first is (a). By , \(F\in\mathcal S^{\perp_{\mathcal D'}}\) and \(\Map(\underline k_{\mathbb A^1},F)\ne0\).