# Wednesday, March 16 ## Grothendieck's Universal Theorem :::{.remark} Setup from last time: $F\in \Add\Cat(\cat A, \cat B)$ left-exact, $\ts{(S^n, \phi_S^n)}_{n\geq 0}$ exact $\delta\dash$functors where for $n > 0$ the $S^n$ are effaceable. Then it is universal: for all $\delta\dash$functors $\ts{(T^n, \phi_T^n)}_{n\geq 0}$ with a natural transformation $S^0 \to T^0$ there exist unique morphisms $(S^n, \phi_S^n) \to (T^n, \phi_T^n)$, i.e. natural transformations $S^n\to T^n$ commuting with the $\phi^n$. ::: ### Proof of Universality :::{.remark} Take an effacement $0\to A \injectsvia{i} M$ for $S^{n+1}$ and extend to a SES $0\to A\to M\to Q\to 0$. We'll define the ladder of morphisms inductively using the following commutative diagram: \begin{tikzcd} {S^nQ} && {S^{n+1}A} && {S^{n+1}M} \\ \\ {T^nQ} && {T^{n+1}A} \arrow["{\phi_S}", from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow["{\phi_T}", from=3-1, to=3-3] \arrow["{f^n}"', from=1-1, to=3-1] \arrow["{\exists f^{n+1} = f^{n+1}(A, i)}", dashed, from=1-3, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJTXm5RIl0sWzIsMCwiU157bisxfUEiXSxbNCwwLCJTXntuKzF9TSJdLFswLDIsIlReblEiXSxbMiwyLCJUXntuKzF9QSJdLFswLDEsIlxccGhpX1MiXSxbMSwyXSxbMyw0LCJcXHBoaV9UIl0sWzAsMywiZl5uIiwyXSxbMSw0LCJcXGV4aXN0cyBmXntuKzF9ID0gZl57bisxfShBLCBpKSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) We need to show - $f^{n+1}(A, i)$ only depends on $A$ - $f^{n+1}$ is functorial in $A$ - $f^{n+1}$ commutes with $\phi_S, \phi_T$. ::: :::{.lemma title="?"} Assume that given two effacements of two delta functors, there exist morphisms: \begin{tikzcd} 0 && {A_1} && {M_1} \\ \\ 0 && {A_2} && {M_2} \arrow[from=1-1, to=1-3, "{g}"] \arrow["{i_1}", from=1-3, to=1-5] \arrow[from=3-1, to=3-3] \arrow[from=1-3, to=3-3] \arrow["{i_2}", from=3-3, to=3-5] \arrow[from=1-5, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCIwIl0sWzIsMCwiQV8xIl0sWzQsMCwiTV8xIl0sWzAsMiwiMCJdLFsyLDIsIkFfMiJdLFs0LDIsIk1fMiJdLFswLDFdLFsxLDIsImlfMSJdLFszLDRdLFsxLDRdLFs0LDUsImlfMiJdLFsyLDVdXQ==) Then there is a commuting square \begin{tikzcd} {S^{n+1}A_1} && {S^{n+1}A_2} \\ \\ {T^{n+1}A_1} && {T^{n+1}A_2} \arrow["{S^{n+1}(g)}", from=1-1, to=1-3] \arrow["{T^{n+1}(g)}", from=3-1, to=3-3] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=3-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJTXntuKzF9QV8xIl0sWzIsMCwiU157bisxfUFfMiJdLFswLDIsIlRee24rMX1BXzEiXSxbMiwyLCJUXntuKzF9QV8yIl0sWzAsMSwiU157bisxfShnKSJdLFsyLDMsIlRee24rMX0oZykiXSxbMSwzXSxbMCwyXV0=) ::: :::{.proof title="?"} There is a cube: \begin{tikzcd} &&& {S^nQ_1} && {S^nQ_2} \\ \\ {S^{n+1}A_1} && {S^{n+1}A_2} & {T^nQ_1} && {T^nQ_2} \\ \\ {T^{n+1}A_1} && {T^{n+1}A_2} \arrow["{S^{n+1}(g)}"{description}, color={rgb,255:red,214;green,92;blue,92}, from=3-1, to=3-3] \arrow["{T^{n+1}(g)}"{description}, color={rgb,255:red,92;green,92;blue,214}, from=5-1, to=5-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, from=3-3, to=5-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-1, to=5-1] \arrow[from=1-4, to=1-6] \arrow[from=3-4, to=3-6] \arrow["{f^{n+1}}"{description}, from=1-6, to=3-6] \arrow["{f^n}"{description}, from=1-4, to=3-4] \arrow["{\phi_S^n}"{description}, color={rgb,255:red,214;green,92;blue,92}, from=1-4, to=3-1] \arrow["{\phi_S^n}"{description}, from=1-6, to=3-3] \arrow["{\phi_T^n}"{description}, from=3-6, to=5-3] \arrow["{\phi_T^n}"{description}, from=3-4, to=5-1] \arrow[from=3-4, to=3-6] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Here all faces but the front form commuting squares. :::{.exercise title="?"} Show that one can move the red path to the blue through the other commuting faces. ::: ::: :::{.corollary title="?"} $f^{n+1}(A, i)$ only depends on $A$. Take two effacements, and assume there is a commuting diagram: \begin{tikzcd} 0 && A && {M_1} \\ \\ 0 && A && {M_2} \arrow[from=1-1, to=1-3] \arrow["{i_1}", from=1-3, to=1-5] \arrow[from=3-1, to=3-3] \arrow["{i_2}", from=3-3, to=3-5] \arrow["{\id_A}"{description}, Rightarrow, no head, from=1-3, to=3-3] \arrow["\exists"{description}, dashed, from=1-5, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCIwIl0sWzIsMCwiQSJdLFs0LDAsIk1fMSJdLFs0LDIsIk1fMiJdLFsyLDIsIkEiXSxbMCwyLCIwIl0sWzAsMV0sWzEsMiwiaV8xIl0sWzUsNF0sWzQsMywiaV8yIl0sWzEsNCwiXFxpZF9BIiwxLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCJcXGV4aXN0cyIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) By the lemma: \begin{tikzcd} {S^{n+1}A} && {S^{n+1}A} \\ \\ {T^{n+1}A} && {T^{n+1}A} \arrow[Rightarrow, no head, from=1-1, to=1-3] \arrow["{f^n(i_1)}"', from=1-1, to=3-1] \arrow[Rightarrow, no head, from=3-1, to=3-3] \arrow["{f^{n}(i_2)}", from=1-3, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJTXntuKzF9QSJdLFsyLDAsIlNee24rMX1BIl0sWzAsMiwiVF57bisxfUEiXSxbMiwyLCJUXntuKzF9QSJdLFswLDEsIiIsMCx7ImxldmVsIjoyLCJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMiwiZl5uKGlfMSkiLDJdLFsyLDMsIiIsMix7ImxldmVsIjoyLCJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzEsMywiZl57bn0oaV8yKSJdXQ==) \begin{tikzcd} 0 && A && {M_1} \\ \\ 0 && A && {M_1\oplus M_2} \\ \\ 0 && A && {M_2} \arrow[from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow[from=3-1, to=3-3] \arrow["{i_2}", from=3-3, to=3-5] \arrow[from=1-1, to=1-3] \arrow["{i_1 \oplus i_2}", from=1-3, to=1-5] \arrow[Rightarrow, no head, from=1-3, to=3-3] \arrow[Rightarrow, no head, from=3-3, to=5-3] \arrow["{(\id_{M_1}, 0)}"', from=3-5, to=1-5] \arrow["{(0, \id_{M_2})}", from=3-5, to=5-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMCwwLCIwIl0sWzIsMCwiQSJdLFs0LDAsIk1fMSJdLFs0LDIsIk1fMVxcb3BsdXMgTV8yIl0sWzQsNCwiTV8yIl0sWzIsMiwiQSJdLFsyLDQsIkEiXSxbMCw0LCIwIl0sWzAsMiwiMCJdLFs3LDZdLFs2LDRdLFs4LDVdLFs1LDMsImlfMiJdLFswLDFdLFsxLDIsImlfMSJdLFsxLDUsIiIsMSx7ImxldmVsIjoyLCJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzUsNiwiIiwxLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMywyLCIoXFxpZF97TV8xfSwgMCkiLDJdLFszLDQsIigwLCBcXGlkX3tNXzJ9KSJdXQ==) ::: > See notes for finished proof.