早期Recollement来自环模挠理论的TTF理论, 即Torsino-Torsion-Free理论.上个世纪80年代, Macpherson和Vilonen将此方法引入一般的Abel范畴中, 而Beilinson, Bernstein, Deligne将此引入三角范畴, 以刻画拓扑空间层上导出范畴.此后, Recollement方法在多个数学分支上获得了许多应用, 如奇异空间几何理论, 表示理论, 多项式算子理论等都有其重要应用^[1].

Recollement的比较函子是研究Recollement的重要概念. Parshall和Scott证明了三角范畴上比较函子都是等价函子, Franjou和Pirashvili在文[5]中通过一个反例说明这些结果对于Abel范畴上的Recollement比较函子并不正确.林亚南和辛林在文[9]中通过倾斜模构造出一类Abel范畴上的非等价的Recollement比较函子.因此Abel范畴上的Recollement比较函子是值得研究的一个问题.

本文将研究由Abel范畴上Recollement的伴随函子诱导的广义Comma范畴上的Recollement, 以及构造由比较函子诱导的新的Recollement.

下面关于Abel范畴上Recollement的定义来自文[9].

定义1.1  设 ${\mathscr C}', {\mathscr{C}}$和 ${\mathscr{C}}^{\prime\prime}$是三个Abel范畴.则范畴 $\mathscr{C}$关于 ${\mathscr{C}}'$和 ${\mathscr{C}}^{\prime\prime}$的Recollement, 记作

是指存在六个正合函子 $i^\ast, i^!:\mathscr{C}\rightarrow\mathscr{C^{\prime}}; i_\ast:\mathscr{C^{\prime}}\rightarrow\mathscr{C}, j_!, j_\ast:\mathscr{C^{\prime\prime}}\rightarrow\mathscr{C}, j^\ast:\mathscr{C}\rightarrow\mathscr{C^{\prime\prime}}$, 并且满足如下三个条件:

(1) $(i^\ast, i_\ast, i^!)$和 $(j_!, j^\ast, j_\ast)$都是伴随三元组;

(2) $i_*, j_!$和 $j_*$都是满且忠实函子;

(3) Im $(i_*)=$ Ker $(j^*)$.

有关Recollement详细内容可参阅文献[2, 3, 5, 7-11].

下面定义属于文[13].

定义1.2  设 $\mathscr{E, C, D}$是范畴, $T: \mathscr{E}\rightarrow \mathscr{C}, H: \mathscr{D}\rightarrow \mathscr{C}$都是函子.构造范畴 $(T\downarrow H)$如下:

(1) $(T\downarrow H)$的对象是所有形如 $\langle E, D, f\rangle $的三元组, 其中 $E\in\rm{Obj}\mathscr{E}$, $D\in\rm{Obj}\mathscr{D}$, $f: T(E)\rightarrow H(D)$是 $\mathscr{C}$中的态射;

(2) 从对象 $\langle E, D, f\rangle $到对象 $\langle E', D', f'\rangle $的态射是所有形如 $\langle e, d\rangle $的态射组, 其中 $e: E\rightarrow E'$是 $\mathscr{E}$中态射, $d: D\rightarrow D'$是 $\mathscr{D}$中态射, 使得 $f'T(e)=H(d)f$, 即有下面的交换图

(3) 态射的合成是分别按 $\mathscr{E}$和 $\mathscr{D}$中态射合成, 称这个范畴 $(T\downarrow H)$是由函子 $T, H$确定的广义Comma范畴.

广义Comma范畴是许多熟悉范畴的自然推广.例如, 设 $\mathscr{E}=\mathscr{C}$是范畴, $B\in\rm{Obj}\mathscr{C}$.如果常值函子 $T_B:\mathscr {E}\rightarrow\mathscr {C}$使得对每一个 $\mathscr {E}$的对象 $M$, $T_B(M)=B$, 对每一个态射 $f$, $T_B(f)=1_B$, 那么广义Comma范畴 $(T_B\downarrow H)$就是通常的Comma范畴 $H\mathscr{D}^B$, 其中范畴 $H\mathscr{D}^B$的对象是 $B\stackrel{f}\rightarrow H(D), \forall D\in\rm{Obj}\mathscr{D}$, 而从 $B\stackrel{f}\rightarrow H(D)$到 $B\stackrel{f^\prime}\rightarrow H(D^\prime)$的态射 $d: f\rightarrow f^{\prime}$是 $\mathscr{D}$中的态射 $d:D\rightarrow D^\prime$使得下图交换

特别地, 当 $\mathscr{E}=\mathscr{C}=\mathscr{D}, H=1_\mathscr{C}$, 范畴 $(T\downarrow H)$是Comma范畴 $\mathscr{C}^B$.

设 $\mathscr{D}=\mathscr{C}$是范畴, $D\in\mathscr{C}$.如果常值函子 $H_D:\mathscr{D}\rightarrow\mathscr{C}$使得对每一个 $\mathscr {D}$的对象 $M$, $H_D(M)=D$, 对每一个态射 $f$, $H_D(f)=1_D$, 那么广义Comma范畴 $(T\downarrow H_D)$是余Comma范畴 $T\mathscr{E}_D$, 其中 $T\mathscr{E}_D$的对象 $ T(E)\stackrel{f}\rightarrow D, \forall E\in\rm{Obj}\mathscr{E}$, 而从 $ T(E)\stackrel{f}\rightarrow D$到 $ T(E^\prime)\stackrel{f^{\prime}}\rightarrow D$的态射 $e: f\rightarrow f^{\prime}$是 $\mathscr{E}$中的态射 $e:E\rightarrow E^\prime$, 使得有下面交换图

特别地, 如果 $\mathscr{D}=\mathscr{C}=\mathscr{E}, T=1_\mathscr{C}$, 那么范畴 $(T\downarrow H)$是余Comma范畴 $\mathscr{C}_D$.

如果 $\mathscr{E}=\mathscr{C}=\mathscr{D}$和 $H=T=1_{\mathscr{C}}$, 那么范畴 $(T\downarrow H)=\mathscr{C}^2$, 这是 $\mathscr{C}$的态射范畴, 其对象是 $\mathscr{C}$中态射, 从对象 $f: A\rightarrow B$到对象 $g: C\rightarrow D$的态射是所有满足下面交换图的 $\mathscr{C}$中态射对 $(t, h)$

如果 $\mathscr{D}=\mathscr{E}$, $\eta: T\rightarrow H$是自然变换, 那么可以构造自然变换范畴 $Nat(\eta)$, 这是 $(T\downarrow H)$的满子范畴, 其对象是 $\langle E, E, \eta_E\rangle, \forall E\in\rm{Obj}\mathscr{E}$, 从 $\langle E, E, \eta_E\rangle $到 $\langle E^\prime, E^\prime, \eta_{E^\prime}\rangle $态射是 $\mathscr{C}$中的态射对 $(e, e)$, 使得有下面的交换图

下面来回忆一下伴随函子.函子 $F:\mathscr{C}\rightarrow \mathscr{D}$和 $G:\mathscr{D}\rightarrow \mathscr{C}$称为伴随的, 如果存在一个自然同构 $\eta=\eta_{X, Y}: \rm{Hom}_{\mathscr{D}}(\mathit{F}(\mathit{X}), \mathit{Y})\rightarrow\rm{Hom}_{\mathscr{C}}(\mathit{X}, \mathit{G}(\mathit{Y}))$, $\forall X\in\rm{Obj}\mathscr{C}, \mathit{Y}\in\rm{Obj}\mathscr{D}$.这时, 也称 $F$是 $G$的左伴随, 而 $G$称为 $F$的右伴随, 或者简称 $(F, G)$是伴随对.下面两个引理可在文[4, 6]中找到.

引理1.3  设函子 $F: \mathscr{A}\rightarrow\mathscr{B}, G: \mathscr{B}\rightarrow\mathscr{A}$构成伴随对 $(F, G)$,

$ h=h_{A, B}: \rm{Hom}_{\mathscr{B}}(\mathit{F}(\mathit{A}), \mathit{B})\rightarrow \rm{Hom}_{\mathscr{A}}(\mathit{A}, \mathit{G}(\mathit{B})) $

是自然同构.那么

(1) 存在自然变换 $\epsilon: 1_{\mathscr{A}}\rightarrow GF$, 使得对每一个 $f\in\rm{Hom}_{\mathscr{B}}(\mathit{F}(\mathit{A}), \mathit{B})$, $h(f)=G(f)\epsilon_{A}$, $\epsilon$也称为该伴随对的单位.

(2) 存在自然变换 $\delta: FG\rightarrow 1_{\mathscr{B}}$使得对每一个 $g\in\rm{Hom}_{\mathscr{A}}(\mathit{A}, \mathit{G}(\mathit{B}))$, $h^{-1}(g)=\delta_{B}F(g)$, $\delta$也称为该伴随对的余单位.

引理1.4  如果 $(F, G)$是伴随对, 那么 $F$是满且忠实函子当且仅当存在自然等价 $GF\approx id$.

定义1.5  设 $\mathscr{E}_1, \mathscr{E}_2, \mathscr{C}_1, \mathscr{C}_2, \mathscr{D}_1, \mathscr{D}_2$都是Abel范畴, $T_1: \mathscr{E}_1\rightarrow\mathscr{C}_1, T_2: \mathscr{E}_2\rightarrow\mathscr{C}_2, H_1: \mathscr{D}_1\rightarrow\mathscr{C}_1, H_2: \mathscr{D}_2\rightarrow\mathscr{C}_2, F_{\mathscr{E}}: \mathscr{E}_1\rightarrow\mathscr{E}_2, F_{\mathscr{C}}: \mathscr{C}_1\rightarrow\mathscr{C}_2, F_{\mathscr{D}}: \mathscr{D}_1\rightarrow\mathscr{D}_2, $ $G_{\mathscr{E}}: \mathscr{E}_2\rightarrow\mathscr{E}_1, G_{\mathscr{C}}: \mathscr{C}_2\rightarrow\mathscr{C}_1, G_{\mathscr{D}}: \mathscr{D}_2\rightarrow\mathscr{D}_1$都是加法函子, 如果 $(F_{\mathscr{E}}, G_{\mathscr{E}})$是伴随对, $\epsilon^{\mathscr{E}_1}, \delta^{\mathscr{E}_2}$分别是该伴随对相应的单位和余单位, $(F_{\mathscr{C}}, G_{\mathscr{C}})$是伴随对, $\epsilon^{\mathscr{C}_1}, \delta^{\mathscr{C}_2}$分别是该伴随对相应的单位和余单位, $(F_{\mathscr{D}}, G_{\mathscr{D}})$是伴随对, $\epsilon^{\mathscr{D}_1}, \delta^{\mathscr{D}_2}$分别是该伴随对相应的单位和余单位, 并且 $F_{\mathscr{C}}T_1=T_2F_{\mathscr{E}}, H_2F_{\mathscr{D}}=F_{\mathscr{C}}H_1, G_{\mathscr{C}}T_2=T_1G_{\mathscr{E}}, H_1G_{\mathscr{D}}=G_{\mathscr{C}}H_2, \epsilon_{T_1E_1}^{\mathscr{C}_1}=T_1(\epsilon_{E_1}^{\mathscr{E}_1}), $$ \delta_{T_2E_2}^{\mathscr{C}_2}=T_2(\delta_{E_2}^{\mathscr{E}_2}), \epsilon_{H_1D_1}^{\mathscr{C}_1}=H_1(\epsilon_{D_1}^{\mathscr{D}_1}), \delta_{H_2D_2}^{\mathscr{C}_2}=H_2(\delta_{D_2}^{\mathscr{D}_2})$, $\forall E_1\in\rm{Obj}\mathscr{E}_1, \mathit{E}_2\in\rm{Obj}\mathscr{E}_2, \mathit{D}_1\in\rm{Obj}\mathscr{D}_1, \mathit{D}_2\in\rm{Obj}\mathscr{D}_2$, 即存在下面交换图

那么, 称 $(F_{\mathscr{E}}, G_{\mathscr{E}}), (F_{\mathscr{C}}, G_{\mathscr{C}}), (F_{\mathscr{D}}, G_{\mathscr{D}})$是相容于 $T_1, T_2, H_1, H_2$的伴随对.

命题2.1  如果函子 $ G: \mathscr{D}\rightarrow\mathscr{C}$有左伴随 $F: \mathscr{C}\rightarrow\mathscr{D}$, 那么

$ (F\downarrow 1_{\mathscr{D}})\cong(1_{\mathscr{C}}\downarrow G). $

证  设 $\epsilon: 1_{\mathscr{C}}\rightarrow GF$是伴随对 $(F, G)$的单位.定义 $Q: (F\downarrow 1_{\mathscr{D}})\rightarrow (1_{\mathscr{C}}\downarrow G)$如下:

(1) $Q:\rm{Obj}(\mathit{F}\downarrow 1_{\mathscr{D}})\rightarrow\rm{Obj}(1_{\mathscr{C}}\downarrow \mathit{G}): \langle \mathit{C}, \mathit{D}, \mathit{f}\rangle \mapsto\langle \mathit{C}, \mathit{D}, \mathit{G}(\mathit{f})\epsilon_\mathit{C}\rangle $, 其中

$ f: F(C)\rightarrow D, G(f)\epsilon_C: C\rightarrow G(D). $

(2) 对任意 $\langle C_1, D_1, f_1\rangle, \langle C_2, D_2, f_2\rangle \in\rm{Obj}(\mathit{F}\downarrow 1_{\mathscr{D}})$, 令

$ \begin{eqnarray*} Q: &&\rm{Hom}_{(\mathit{F}\downarrow 1_{\mathscr{D}})}(\langle \mathit{C}_1, \mathit{D}_1, \mathit{f}_1\rangle, \langle \mathit{C}_2, \mathit{D}_2, \mathit{f}_2\rangle )\\ &\rightarrow&\rm{Hom}_{(1_{\mathscr{C}}\downarrow \mathit{G})}(\mathit{Q}\langle \mathit{C}_1, \mathit{D}_1, \mathit{f}_1\rangle, \mathit{Q}\langle \mathit{C}_2, \mathit{D}_2, \mathit{f}_2\rangle ):\langle \mathit{a}, \mathit{b}\rangle \mapsto\langle \mathit{a}, \mathit{b}\rangle .\end{eqnarray*} $

由于 $bf_1=f_2F(a)$, 则

$ G(b)G(f_1)\epsilon_{C_1}=G(f_2F(a))\epsilon_{C_1}=G(f_2)GF(a)\epsilon_{C_1}=G(f_2) \epsilon_{C_2}a, $

所以 $Q$是定义好的.容易知道 $Q$是忠实的.

对 $\langle a, b\rangle : \langle C_1, D_1, G(f_1)\epsilon_{C_1}\rangle \rightarrow {\langle C_2, D_2, G(f_2)\epsilon_{C_2}\rangle }\in\rm{Mor}{(1_{\mathscr{C}}\downarrow \mathit{G})}, $

$ G(b)G(f_1)\epsilon_{C_1}=G(f_2)\epsilon_{C_2}a, $

故

$ \begin{eqnarray*} bf_1&=&b\delta_{D_1}F(G(f_1)\epsilon_{C_1})=\delta_{D_2}FG(b)F(G(f_1)\epsilon_{C_1})\\ &=&\delta_{D_2}F(G(f_2)\epsilon_{C_2}a)=\delta_{D_2}F(G(f_2)\epsilon_{C_2})F(a)\\ &=&f_2F(a). \end{eqnarray*} $

因此 $Q\langle a, b\rangle =\langle a, b\rangle $, 即 $Q$是满的. $Q$显然是稠密的.所以 $(F\downarrow 1_{\mathscr{D}})\cong(1_{\mathscr{C}}\downarrow G)$.

下面结论类似可得.

命题2.2  如果 $F: \mathscr{C}\rightarrow\mathscr{D}, G: \mathscr{D}\rightarrow\mathscr{C}$是函子使得 $(F, G)$是伴随对, 那么

$ (F\downarrow FG)\cong (1_{\mathscr{C}}\downarrow GFG), (G\downarrow GF)\cong (FG\downarrow F), \\ (F\downarrow F)\cong (1_{\mathscr{C}}\downarrow GF), (G\downarrow G)\cong (FG\downarrow 1_{\mathscr{D}}). $

推论2.3  如果下图

是 $\mathscr{C}$关于 $\mathscr{C}^\prime, \mathscr{C}^{\prime\prime}$的Recollement, 那么

(1) $\begin{array}{cc}(i^\ast\downarrow 1_{\mathscr{C^\prime}})\cong(1_{\mathscr{C}}\downarrow i_\ast),& (i_\ast\downarrow 1_{\mathscr{C}})\cong(1_{\mathscr{C^\prime}}\downarrow i^!), \\ (j_!\downarrow 1_{\mathscr{C}})\cong (1_{\mathscr{C^{\prime\prime}}}\downarrow j^\ast),&(j^\ast\downarrow 1_{\mathscr{C^{\prime\prime}}})\cong (1_{\mathscr{C}}\downarrow j_*).\end{array}$

(2) $(i_\ast\downarrow i_\ast i^!)\cong(1_{\mathscr{C^\prime}}\downarrow i^!), \, \, \, (j_!\downarrow j_!j^*)\cong(1_{\mathscr{C^{\prime\prime}}}\downarrow j^\ast)$.

(3) $\begin{array}{cc}(i_*\downarrow i_*i^*)\cong(1_{\mathscr{C^\prime}}\downarrow i^*),& (i^!\downarrow 1_{\mathscr{C^\prime}})\cong(i_\ast i^!\downarrow i_\ast), \\ (j^\ast\downarrow 1_{\mathscr{C^{\prime\prime}}})\cong(j_!j^\ast\downarrow j_!),&(j_\ast\downarrow j_\ast j^\ast)\cong(1_{\mathscr{C^{\prime\prime}}}\downarrow j^\ast).\end{array}$

(4) $(i^\ast\downarrow i^\ast)\cong(1_{\mathscr{C}}\downarrow i_\ast i^\ast), (i_\ast\downarrow i_\ast)\cong {\mathscr{C^\prime}}^2, (j_!\downarrow j_!)\cong {\mathscr{C^{\prime\prime}}^2}, (j^\ast\downarrow j^\ast)\cong (1_{\mathscr{C}}\downarrow j_\ast j^\ast)$.

(5) $(i^!\downarrow i^!)\cong(i_*i^!\downarrow 1_{\mathscr{C}}), (j^\ast\downarrow j^\ast)\cong(j_!j^\ast\downarrow 1_{\mathscr{C}}), (j_\ast\downarrow j_\ast)\cong {\mathscr{C^{\prime\prime}}^2}$.

命题2.4  设 $T: \mathscr{E}\rightarrow \mathscr{C}$和 $H: \mathscr{D}\rightarrow \mathscr{C}$是函子.如果对所有的 $E\in\rm{Obj}{\mathscr{E}}, \mathit{D}\in\rm{Obj}{\mathscr{D}}$, $\rm{Hom}_{\mathscr{C}}(\mathit{T}(\mathit{E}), \mathit{H}(\mathit{D}))=0$, 那么 $(T\downarrow H)\cong\mathscr{E}\times\mathscr{D}$.

证  设 $F:\mathscr{E}\times\mathscr{D}\rightarrow(T\downarrow H)$使得 $F(E, D)= \langle E, D, 0\rangle .$那么 $F$是范畴同构.

推论2.5  如果下图

是 $\mathscr{C}$关于 $\mathscr{C}^\prime, \mathscr{C}^{\prime\prime}$的Recollement, 那么

(1) $(j_!\downarrow i_\ast)\cong \mathscr{C^{\prime\prime}}\times\mathscr{C^\prime}, (j_!j^\ast\downarrow i_\ast)\cong \mathscr{C}\times\mathscr{C^\prime}, (j_!\downarrow i_\ast i^\ast)\cong (j_!\downarrow i_\ast i^!)\cong \mathscr{C^{\prime\prime}}\times\mathscr{C}.$

(2) $(i_\ast\downarrow j_\ast)\cong\mathscr{C^\prime}\times\mathscr{C^{\prime\prime}}, (i_\ast i^!\downarrow j_\ast)\cong(i_\ast i^\ast\downarrow j_\ast)\cong\mathscr{C}\times\mathscr{C^{\prime\prime}}, (i_\ast\downarrow j_\ast j^\ast)\cong\mathscr{C^\prime}\times\mathscr{C}.$

(3) $(j_!j^\ast\downarrow i_\ast i^\ast)\cong (j_!j^\ast\downarrow i_\ast i^!)\cong(i_\ast i^!\downarrow j_\ast j^\ast)\cong(i_\ast i^\ast\downarrow j_\ast j^\ast)\cong\mathscr{C}\times\mathscr{C}.$

引理3.1  设 $\mathscr{E}, \mathscr{D}$和 $\mathscr{C}$都是Abel范畴.如果 $(H, T)$是伴随对, 并且 $T: \mathscr{E}\rightarrow \mathscr{C}$是右正合函子, $H: \mathscr{D}\rightarrow \mathscr{C}$是左正合函子, 则 $(T\downarrow H)$是Abel范畴.

证  由于 $T, H$是加法函子, 所以 $(T\downarrow H)$是预加法范畴.

设 $E_i\in\rm{Obj}{\mathscr{E}}, \mathit{D_i}\in\rm{Obj}{\mathscr{D}}, \mathit{i}\in \mathit{I}$, 其中 $I$是有限集合.令 $E=\amalg E_i$和 $D=\amalg D_i$分别是这些对象的上积.由于加法函子保持有限上积, 故 $T(E)=\amalg T(E_i)$和 $H(D)=\amalg H(D_i)$.于是 $\langle E, D, f\rangle =\amalg \langle E_i, D_i, f_i\rangle $, 其中 $f=\amalg f_i$.事实上, 如果 $e_i: E_i\rightarrow E, d_i: D_i\rightarrow D$是典范的嵌入态射, 则 $\{\langle e_i, d_i\rangle : {\langle E_i, D_i, f_i\rangle }\rightarrow \langle E, D, f\rangle \}_{i\in I}$是 $(T\downarrow H)$中的一类态射.对每一个 ${\langle M, N, k\rangle }\in\rm{Obj}(\mathit{T}\downarrow \mathit{H})$, 如果存在 $\langle m_i, n_i\rangle : {\langle E_i, D_i, f_i\rangle }\rightarrow {\langle M, N, k\rangle }$, 则存在 $m: E\rightarrow M, n: D\rightarrow N$使得 $m_i=me_i, n_i=nd_i$, 所以

$ kT(m)T(e_i)=kT(m_i)=H(n_i)f_i=H(n)H(d_i)f_i=H(n)fT(e_i). $

故 $kT(m)=H(n)f$.因此 $\langle m, n\rangle \in\rm{Hom}_{(\mathit{T}\downarrow \mathit{H})}(\langle \mathit{E}, \mathit{D}, \mathit{f}\rangle, \langle \mathit{M}, \mathit{N}, \mathit{k}\rangle )$, 这表明 $(T\downarrow H)$是加法范畴.

对任意 $\langle e, d\rangle \in \rm{Hom}_{(\mathit{T}\downarrow \mathit{H})}(\langle \mathit{E}_1, \mathit{D}_1, \mathit{f}_1\rangle, \langle \mathit{E}_2, \mathit{D}_2, \mathit{f}_2\rangle )$, 存在 $\rm{ker}\mathit{e}: \rm{Ker}\mathit{e}\rightarrow \mathit{E}_1$和 $\rm{ker}\mathit{d}: \rm{Ker}\mathit{d}\rightarrow \mathit{D}_1$.由于 $H$是左正合函子, 从而保持核, 于是存在唯一态射 $f_0: T(\rm{Ker}\mathit{e})\rightarrow \mathit{H}(\rm{Ker}\mathit{d})$使得下图交换

故

$ \langle \rm{ker}\mathit{e}, \rm{ker}\mathit{d}\rangle \in\rm{Hom}_{(\mathit{T}\downarrow \mathit{H})}(\langle \rm{ker}\mathit{e}, \rm{ker}\mathit{d}, \mathit{f}_0\rangle, \langle \mathit{E}_1, \mathit{D}_1, \mathit{f}_1\rangle ) $

使得 $\langle e, d\rangle \langle \rm{ker}\mathit{e}, \rm{ker}\mathit{d}\rangle =0.$如果 $\langle M, N, h\rangle \in$ Obj $(T\downarrow H)$和 $\langle a, b\rangle :\langle M, N, h\rangle \rightarrow \langle E_1, D_1, f_1\rangle $使得 $\langle e, d\rangle \langle a, b\rangle =0$, 那么存在唯一 $a_1: M\rightarrow \rm{ker}\mathit{e}$和 $b_1:N\rightarrow\rm{ker}\mathit{d}$使得 $a=\rm{ker}\mathit{e}\cdot a_1, b=\rm{ker}\mathit{d}\cdot b_1$.故 $\mathit{T}(\mathit{a})=\mathit{T}(\rm{ker}\mathit{e})\cdot \mathit{T}(\mathit{a}_1), \mathit{H}(\mathit{b})=\mathit{H}(\rm{ker}\mathit{d})\cdot \mathit{H}(\mathit{b}_1)$, 从而由 $H(\rm{ker}\mathit{d})(\mathit{f}_0\mathit{T}(\mathit{a}_1)-\mathit{H}(\mathit{b}_1)\mathit{h})=0$得 $f_0T(a_1)=H(b_1)h$.所以 $\langle \rm{ker}\mathit{e}, \rm{ker}\mathit{d}\rangle =\rm{ker}\langle \mathit{e}, \mathit{d}\rangle $.对偶地, 如果 $T$右正合函子, 那么在 $(T\downarrow H)$中每一个态射都有余核.由于 $\mathscr{E}, \mathscr{D}$都是Abel范畴, 所以在 $(T\downarrow H)$中每一个态射都是严格的.因此 $(T\downarrow H)$是Abel范畴.这证明了引理.

设 $\mathscr{E}_1, \mathscr{E}_2, \mathscr{D}_1, \mathscr{D}_2$是Abel范畴, $\mathscr{C}_1, \mathscr{C}_2$是加法范畴.如果下图是加法函子交换图

那么存在从广义Comma范畴 $(T_1\downarrow H_1)$到广义Comma范畴 $(T_2\downarrow H_2)$的加法函子 $\widetilde{F}$使得

$ \begin{eqnarray*}\widetilde{F}\langle E_1, D_1, f_1\rangle &=&\langle F_{\mathscr{E}}(E_1), F_{\mathscr{D}}(D_1), F_{\mathscr{C}}(f_1)\rangle, \\ \widetilde{F}\langle e, d\rangle &=&\langle F_{\mathscr{E}}(e), F_{\mathscr{D}}(d)\rangle .\end{eqnarray*} $

上述 $\widetilde{F}$称为由 $(F_{\mathscr{E}}, F_{\mathscr{D}})$诱导的函子.

引理3.2  设 $\mathscr{E}_1, \mathscr{E}_2, \mathscr{D}_1, \mathscr{D}_2$是Abel范畴, $\mathscr{C}_1, \mathscr{C}_2$是加法范畴.在下面的交换图

如果 $(F_{\mathscr{E}}, G_{\mathscr{E}}), (F_{\mathscr{C}}, G_{\mathscr{C}}), (F_{\mathscr{D}}, G_{\mathscr{D}})$是相容于 $T_1, T_2, H_1, H_2$的伴随对, 那么存在伴随对 $(\widetilde{F}, \widetilde{G})$, 其中 $\widetilde{F}:(T_1\downarrow H_1)\rightarrow (T_2\downarrow H_2)$和 $\widetilde{G}:(T_2\downarrow H_2)\rightarrow (T_1\downarrow H_1)$分别是由 $(F_{\mathscr{E}}, F_{\mathscr{D}})$和 $(G_{\mathscr{E}}, G_{\mathscr{D}})$诱导的函子.

证  设 $\widetilde{F}:(T_1\downarrow H_1)\rightarrow (T_2\downarrow H_2)$和 $\widetilde{G}:(T_2\downarrow H_2)\rightarrow (T_1\downarrow H_1)$分别是由 $(F_{\mathscr{E}}, F_{\mathscr{D}})$和 $(G_{\mathscr{E}}, G_{\mathscr{D}})$诱导的函子.如果

$ \langle E_1, D_1, f_1\rangle \in\rm{Obj}(\mathit{T}_1\downarrow \mathit{H}_1), \langle \mathit{E}_2, \mathit{D}_2, \mathit{f}_2\rangle \in\rm{Obj}(\mathit{T}_2\downarrow \mathit{H}_2), $

对每一个 $\langle e, d\rangle \in \rm{Hom}_{(\mathit{T}_1\downarrow \mathit{H}_1)}(\langle \mathit{E}_1, \mathit{D}_1, \mathit{f}_1\rangle, \widetilde{G}\langle \mathit{E}_2, \mathit{D}_2, \mathit{f}_2\rangle )$, 由于 $H_1(d)f_1=G_{\mathscr{C}}(f_2)T_1(e)$, 则

$ \begin{eqnarray*} H_2(\delta_{D_2}^{\mathscr{D}}F_{\mathscr{D}}(d)))F_{\mathscr{C}}(f_1)&=&H_2(\delta_{D_2}^{\mathscr{D}})F_{\mathscr{C}}H_1(d)F_{\mathscr{C}}(f_1)\\ &=& H_2(\delta_{D_2}^{\mathscr{D}})F_{\mathscr{C}}(H_1(d)f_1)\\ &=&H_2(\delta_{D_2}^{\mathscr{D}})F_{\mathscr{C}}(G_{\mathscr{C}}(f_2)T_1(e))\\ &=&H_2(\delta_{D_2}^{\mathscr{D}})F_{\mathscr{C}}G_{\mathscr{C}}(f_2)T_2F_{\mathscr{E}}(e)\\ &=&\delta_{H_2D_2}^{\mathscr{C}}F_{\mathscr{C}}G_{\mathscr{C}}(f_2)T_2F_{\mathscr{E}}(e)\\ &=&f_2T_2(\delta_{E_2}^{\mathscr{E}})T_2F_{\mathscr{E}}(e)\\ &=&f_2T_2(\delta_{E_2}^{\mathscr{E}}F_{\mathscr{E}}(e)).\end{eqnarray*} $

故 $(\delta_{E_2}^{\mathscr{E}}F_{\mathscr{E}}(e), \delta_{D_2}^{\mathscr{D}}F_{\mathscr{D}}(d))\in \rm{Hom}_{(\mathit{T}_2\downarrow \mathit{H}_2)}(\widetilde{\mathit{F}}\langle \mathit{E}_1, \mathit{D}_1, \mathit{f}_1\rangle, \langle \mathit{E}_2, \mathit{D}_2, \mathit{f}_2\rangle )$, 所以映射

$ \begin{eqnarray*}\phi:&&\rm{Hom}_{(\mathit{T}_1\downarrow \mathit{H}_1)}(\langle \mathit{E}_1, \mathit{D}_1, \mathit{f}_1\rangle, \widetilde{G}\langle \mathit{E}_2, \mathit{D}_2, \mathit{f}_2\rangle )\\ &\rightarrow &{\rm{Hom}}_{(T_2\downarrow H_2)}(\widetilde{F}\langle E_1, D_1, f_1\rangle, \langle E_2, D_2, f_2\rangle )\end{eqnarray*} $

使得 $\phi(\langle e, d\rangle )=\langle \delta_{E_2}^{\mathscr{E}}F_{\mathscr{E}}(e), \delta_{D_2}^{\mathscr{D}}F_{\mathscr{D}}(d)\rangle $是合理的.类似地, 存在映射

$ \begin{eqnarray*}\psi:&&{\rm{Hom}}_{(T_2\downarrow H_2)}(\widetilde{F}\langle E_1, D_1, f_1\rangle, \langle E_2, D_2, f_2\rangle )\\ &\rightarrow&{\rm{Hom}}_{(T_1\downarrow H_1)}(\langle E_1, D_1, f_1\rangle, \widetilde{G}\langle E_2, D_2, f_2\rangle )\end{eqnarray*} $

使得 $\psi(\langle e', d'\rangle )=\langle G_{\mathscr{E}}(e')\epsilon_{E_1}^{\mathscr{E}}, G_{\mathscr{D}}(d')\epsilon_{D_1}^{\mathscr{D}}\rangle $.而且

$ \begin{eqnarray*} \psi\phi(e, d)&=&\psi(\delta_{E_2}^{\mathscr{E}}F_{\mathscr{E}}(e), \delta_{D_2}^{\mathscr{D}}F_{\mathscr{D}}(d))\\ &=&(G_{\mathscr{E}}(\delta_{E_2}^{\mathscr{E}}F_{\mathscr{E}}(e))\epsilon_{E_1}^{\mathscr{E}}, G_{\mathscr{D}}(\delta_{D_2}^{\mathscr{D}}F_{\mathscr{D}}(d))\epsilon_{D_1}^{\mathscr{D}})\\ &=&(e, d), \end{eqnarray*} $

于是 $\psi\phi=1$.类似 $\phi\psi=1$.所以

$ \begin{eqnarray*}&\rm{Hom}_{(\mathit{T}_1\downarrow \mathit{H}_1)}(\langle \mathit{E}_1, \mathit{D}_1, \mathit{f}_1\rangle, \widetilde{G}\langle \mathit{E}_2, \mathit{D}_2, \mathit{f}_2\rangle )\\ \cong&{\rm{Hom}_{(\mathit{T}_2\downarrow \mathit{H}_2)}(\widetilde{\mathit{F}}\langle \mathit{E}_1, \mathit{D}_1, \mathit{f}_1\rangle, }\langle \mathit{E}_2, \mathit{D}_2, \mathit{f}_2\rangle ).\end{eqnarray*} $

下面将证明 $\phi, \psi$是自然的.

对

$ \begin{eqnarray*}&&\langle E^\prime, D^\prime, f^\prime\rangle \in{\rm{Obj}}(T_2\downarrow H_2), \\ &&\langle a, b\rangle \in {\rm{Hom}}_{(T_1\downarrow H_1)}({\langle E_2, D_2, f_2\rangle, \langle E_1, D_1, f_1\rangle )}, \end{eqnarray*} $

下图是交换的

事实上, 对每一个 $\langle e', d'\rangle \in {\rm{Hom}}_{(T_2\downarrow H_2)}(\widetilde{F}\langle E_1, D_1, f_1\rangle, \langle E', D', f'\rangle )$,

$ \begin{eqnarray*} \psi_2(\widetilde{F}\langle a, b\rangle )^\ast\langle e', d'\rangle &=&\psi_2\langle F_{\mathscr{E}}(a), F_{\mathscr{D}}(b)\rangle ^\ast\langle e', d'\rangle \\ &=&\psi_2\langle e'F_{\mathscr{E}}(a), d'F_{\mathscr{D}}(b)\rangle \\ &=&\langle G_{\mathscr{E}}(e'F_{\mathscr{E}}(a))\epsilon_{E_2}^{\mathscr{E}_1}, G_{\mathscr{D}}(d'F_{\mathscr{D}}(b))\epsilon_{D_2}^{\mathscr{D}_1}\rangle, \\ \langle a, b\rangle ^\ast\psi_1\langle e', d'\rangle &=&\langle a, b\rangle ^\ast\langle G_{\mathscr{E}}(e')\epsilon_{E_1}^{\mathscr{E}_1}, G_{\mathscr{D}}(d')\epsilon_{D_1}^{\mathscr{D}_1}\rangle \\ &=& \langle G_{\mathscr{E}}(e')\epsilon_{E_1}^{\mathscr{E}_1}a, G_{\mathscr{D}}(d')\epsilon_{D_1}^{\mathscr{D}_1}b\rangle .\end{eqnarray*} $

于是

$ \begin{eqnarray*}G_{\mathscr{E}}(e'F_{\mathscr{E}}(a))\epsilon_{E_2}^{\mathscr{E}_1}&=&G_{\mathscr{E}}(e')G_{\mathscr{E}}F_{\mathscr{E}}(a)\epsilon_{E_2}^{\mathscr{E}_1} =G_{\mathscr{E}}(e')\epsilon_{E_1}^{\mathscr{E}_1}a, \\ G_{\mathscr{D}}(d'F_{\mathscr{D}}(b))\epsilon_{D_2}^{\mathscr{D}_1}&=& G_{\mathscr{D}}(d')G_{\mathscr{D}}F_{\mathscr{D}}(b)\epsilon_{D_2}^{\mathscr{D}_1}=G_{\mathscr{D}}(d')\epsilon_{D_1}^{\mathscr{D}_1}b. \end{eqnarray*} $

故 $\psi_2(\widetilde{F}\langle a, b\rangle )^\ast=\langle a, b\rangle ^\ast\psi_1$, 即 $\psi$在 $\langle E_1, D_1, f_1\rangle $处是自然的.类似 $\psi$在 $\langle E_2, D_2, f_2\rangle $处是自然的, $\phi$也是自然的.所以 $(\widetilde{F}, \widetilde{G})$是伴随对.

下面定理是本文的主要结果:

定理3.3  设 $\mathscr{E}_1, \mathscr{E}_2, \mathscr{E}_3, \mathscr{C}_1, \mathscr{C}_2, \mathscr{C}_3, \mathscr{D}_1, \mathscr{D}_2, \mathscr{D}_3$都是Abel范畴.在下面的Recollement交换图中, 如果所有伴随对关于相应函子都是相容的,

$T_1, T_2, T_3$是右正合函子, $H_1, H_2, H_3$是左正合函子, 那么存在广义Comma范畴的Recollement.

证  由引理3.1, $(T_i\downarrow H_i)$都是Abel范畴, $i=1, 2, 3$.根据引理3.2, 分别存在由 $i^\ast, j_!, i_\ast, j^\ast, i^!, j_\ast$诱导的加法函子 $\widetilde{F}^\ast, \widetilde{G}_!, \widetilde{F}_\ast, \widetilde{G}^\ast, \widetilde{F}^!, \widetilde{G}_\ast$.而且 $(\widetilde{F}^\ast, \widetilde{F}_\ast, \widetilde{F}^!), (\widetilde{G}_!, \widetilde{G}^\ast, \widetilde{G}_\ast)$都是伴随三元组.

对任意 $\langle E_1, D_1, f_1\rangle \in{\rm{Obj}}(T_1\downarrow H_1)$,

$ \widetilde{F}^\ast\widetilde{F}_\ast(\langle E_1, D_1, f_1\rangle )=\langle i^\ast_{\mathscr{E}}i_\ast^{\mathscr{E}}E_1, i^\ast_{\mathscr{D}}i_\ast^{\mathscr{D}}D_1, i^\ast_{\mathscr{C}}i_\ast^{\mathscr{C}}f_1\rangle \simeq\\\langle E_1, D_1, f_1\rangle . $

对每一个 $\langle e, d\rangle \in{\rm{Hom}}_{(T_1\downarrow H_1)}(\langle E_1, D_1, f_1\rangle, \langle E_2, D_2, f_2\rangle )$,

$ \widetilde{F}^\ast\widetilde{F}_\ast\langle e, d\rangle =\langle i^\ast_{\mathscr{E}}i_\ast^{\mathscr{E}}(e), i^\ast_{\mathscr{D}}i_\ast^{\mathscr{D}}(d)\rangle \simeq\langle e, d\rangle, $

因此 $\widetilde{F}^\ast\widetilde{F}_\ast\simeq 1_{(T_1\downarrow H_1)}$.类似地, $\widetilde{F}^!\widetilde{F}_\ast\simeq 1_{(T_1\downarrow H_1)}, \widetilde{G}^\ast\widetilde{G}_!\simeq 1_{(T_3\downarrow H_3)}, $ $\widetilde{G}^\ast\widetilde{G}_\ast \simeq 1_{(T_3\downarrow H_3)}$.因此由引理1.4可得 $\widetilde{F}_\ast, \widetilde{G}_\ast$和 $\widetilde{G}_!$都是满且忠实函子.

对每一个 $\langle E_1, D_1, f_1\rangle \in{\rm{Obj}}(T_1\downarrow H_1)$,

$ \widetilde{G}^\ast\widetilde{F}_\ast(\langle E_1, D_1, f_1\rangle )=\langle j^\ast_{\mathscr{E}}i_\ast^{\mathscr{E}}E_1, j^\ast_{\mathscr{D}}i_\ast^{\mathscr{D}}D_1, j^\ast_{\mathscr{C}}i_\ast^{\mathscr{C}}f_1\rangle =0. $

对每一个 $\langle e, d\rangle \in{\rm{Hom}}_{(T_1\downarrow H_1)}(\langle E_1, D_1, f_1\rangle, \langle E_2, D_2, f_2\rangle )$,

$ \widetilde{G}^\ast\widetilde{F}_\ast\langle e, d\rangle =\langle j^\ast_{\mathscr{E}}i_\ast^{\mathscr{E}}(e), j^\ast_{\mathscr{D}}i_\ast^{\mathscr{D}}(d)\rangle =0, $

因此 $\widetilde{G}^\ast\widetilde{F}_\ast=0$.

如果 $\widetilde{G}^\ast:(T_2\downarrow H_2)\rightarrow (T_3\downarrow H_3)$使得 $\widetilde{G}^\ast\langle E_2, D_2, f_2\rangle =\langle j^\ast_{\mathscr{E}}(E_2), j^\ast_{\mathscr{D}}(D_2), j^\ast_{\mathscr{C}}(f_2)\rangle =0$, 那么存在 $\mathscr{E}$的对象 $E_1$和 $\mathscr{D}$的对象 $D_1$使得 $E_2=i^{\mathscr{E}}_\ast(E_1), D_2=i^{\mathscr{D}}_\ast(D_1)$.由于 $i^{\mathscr{C}}_\ast$是满的, 存在 $\mathscr{C}_1$中态射 $f_1:T_1E_1\rightarrow H_1D_1$使得 $i^{\mathscr{C}}_\ast(f_1)=f_2: i^{\mathscr{C}}_\ast T_1E_1= T_2E_2\rightarrow H_2D_2=i^{\mathscr{C}}_\ast H_1D_1$.所以 $\langle E_1, D_1, f_1\rangle \in$ Obj $(T_1\downarrow H_1)$并且

$ \widetilde{F}_\ast\langle E_1, D_1, f_1\rangle =\langle i^{\mathscr{E}}_\ast(E_1), i^{\mathscr{D}}_\ast(D_1), i^{\mathscr{C}}_\ast(f_1)\rangle =\langle E_2, D_2, f_2\rangle . $

定理证毕.

下面考察一个来自文[5]的例子.

例1  设 $\mathscr{C}_1$是2元域 $\mathbb{Z}_2$上有限维向量空间范畴, $\mathscr{C}_2$是2元域 $\mathbb{Z}_2$上具有对合的有限维向量空间范畴.如果 $\mathscr{C}$是2元域 $\mathbb{Z}_2$上有限维向量空间如下形式对象的范畴:

$ (V_1, H, V_2, P): V_1\rightleftarrows V_2, $

其中 $H: V_1\rightarrow V_2, P: V_2\rightarrow V_1$都是线性映射, 并且满足 $PHP = 0$和 $HPH = 0$, 那么 $\mathscr{C}$关于 $\mathscr{C}_1$和 $\mathscr{C}_2$的Recollement如下图所示.

具有如下函子:

$ \begin{eqnarray*}&&i^*(V_1, H, V_2, P)={\rm{Coker}} (P), j_!(V, T)=(V_T, 1+T, V, p), \\ &&i_*(V)=(V, 0, 0, 0), j^*(V_1, H, V_2, P)=(V_2, HP-1), \\ &&i^!(V_1, H, V_2, P)={\rm{Ker}} (H), j_*(V, T)=(V^T, h, V, 1+T), \end{eqnarray*} $

其中 $V^T ={\rm{Ker}}(1-T), V_T ={\rm{Coker}}(1-T)$, $h$是典范的嵌入, $p$是典范的商映射.

现在考察 $\mathscr{C}$的满子范畴 $\mathscr{B}$和 $\mathscr{A}$. $\mathscr{B}$的对象是由所有满足 $PH = 0$的对象 $(V_1, H, V_2, P)$构成, $\mathscr{A}$的对象是由所有满足 $HP=0$的对象 $(V_1, H, V_2, P)$构成.

类似文[5], 可得如下的Recollements.

于是包含函子 $L: \mathscr{B}\rightarrow\mathscr{C}$和 $F: \mathscr{A}\rightarrow\mathscr{C}$分别给出了相应的比较函子.由于 $\mathbb{Z}_2$上有限维向量空间范畴上的嵌入函子都是正合函子, 因此 $L, F$均为正合函子.

存在下面的交换图

根据定理3.3, 可得广义Comma范畴的Recollement.

由于 $\mathscr{C}_1^2\cong (1_{\mathscr{C}_1}\downarrow 1_{\mathscr{C}_1}), \mathscr{C}_2^2\cong (1_{\mathscr{C}_2}\downarrow 1_{\mathscr{C}_2})$, 所以有下面的Recollement.