# Monday, March 01 :::{.remark} Last time: we looked at group extensions. Given $\xi: 0\to B\to X \to A\to 0$, we had a canonical element in $\Ext^1(A, B)$, namely $\Theta(\xi) = \delta(\one_B)$. This only depends on the equivalence class of $\xi$. ::: :::{.theorem title="Module extensions biject with Ext groups"} Given $A, B\in \modr$, there is a bijection \[ \correspond{ \text{Extensions of $A$ by $B$} } \mapscorrespond{\Phi}{\Theta} \Ext_R^1(A, B) \] ::: :::{.proof title="?"} \envlist :::{.claim} $\Theta$ is surjective. ::: Fix a SES \[ 0 \to M \mapsvia{j} P \mapsvia{\pi} A \to 0 \] with $P$ projective, and take the LES resulting from applying $\Hom(\wait, B)$: \begin{tikzcd} 0 \\ {\Hom(A, B)} & {\Hom(P, B)} & {\Hom(M, B)} \\ \\ {\Ext^1(A, B)} & {\Ext^1(P, B) = 0} \\ x \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow["\bd", from=2-3, to=4-1, out=0, in=180] \arrow[from=4-1, to=4-2] \arrow[from=1-1, to=2-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwxLCJcXEhvbShBLCBCKSJdLFsxLDEsIlxcSG9tKFAsIEIpIl0sWzIsMSwiXFxIb20oTSwgQikiXSxbMCwzLCJcXEV4dF4xKEEsIEIpIl0sWzEsMywiXFxFeHReMShQLCBCKSA9IDAiXSxbMCw0LCJ4Il0sWzAsMCwiMCJdLFswLDFdLFsxLDJdLFsyLDMsIlxcYmQiXSxbMyw0XSxbNiwwXV0=) Letting $x \in \Ext^1(A, B)$ and choose $\beta\in \Hom(M, B)$ with $\bd \beta = x$ using that $P$ is projective and thus $\Ext^1(P, B)$ vanishes. Now let $X$ be the **pushout** of $j: M\to P$ and $\beta: M\to B$. Note that we can apply the universal property of cokernels to get a map of the following form: \begin{tikzcd} M && {P\oplus B} && {X = \coker g} && 0 \\ \\ && A \arrow["{g = (j, -\beta)}", from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow["{\pi \oplus 0}", from=1-3, to=3-3] \arrow[from=1-5, to=1-7] \arrow["{\therefore 0}"', dotted, from=1-1, to=3-3] \arrow["{\exists ! \mu}", dashed, from=1-5, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJNIl0sWzIsMCwiUFxcb3BsdXMgQiJdLFs0LDAsIlggPSBcXGNva2VyIGciXSxbMiwyLCJBIl0sWzYsMCwiMCJdLFswLDEsImcgPSAoaiwgLVxcYmV0YSkiXSxbMSwyXSxbMSwzLCJcXHBpIFxcb3BsdXMgMCJdLFsyLDRdLFswLDMsIlxcdGhlcmVmb3JlIDAiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkb3R0ZWQifX19XSxbMiwzLCJcXGV4aXN0cyAhIFxcbXUiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) Taking the pushout yields a diagram: \begin{tikzcd} 0 && M && P && A && 0 \\ \\ 0 && B && X && A && 0 \arrow[from=1-1, to=1-3] \arrow["j"', from=1-3, to=1-5] \arrow["\pi"', from=1-5, to=1-7] \arrow[from=3-1, to=3-3] \arrow["\iota"', from=3-3, to=3-5] \arrow["\mu"', from=3-5, to=3-7] \arrow[from=1-1, to=3-1] \arrow["\beta"', from=1-3, to=3-3] \arrow["\sigma"{description}, from=1-5, to=3-5] \arrow[equals, from=1-7, to=3-7] \arrow[from=1-7, to=1-9] \arrow[from=3-7, to=3-9] \arrow[from=1-9, to=3-9] \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=180}, draw=none, from=3-5, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFswLDIsIjAiXSxbMiwwLCJNIl0sWzIsMiwiQiJdLFs0LDIsIlgiXSxbNCwwLCJQIl0sWzYsMCwiQSJdLFs2LDIsIkEiXSxbOCwwLCIwIl0sWzgsMiwiMCJdLFswLDJdLFsyLDUsImoiLDJdLFs1LDYsIlxccGkiLDJdLFsxLDNdLFszLDQsIlxcaW90YSIsMl0sWzQsNywiXFxtdSIsMl0sWzAsMV0sWzIsMywiXFxiZXRhIiwyXSxbNSw0LCJcXHNpZ21hIiwxXSxbNiw3LCIiLDEseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzYsOF0sWzcsOV0sWzgsOV0sWzQsMiwiIiwyLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d) :::{.exercise title="?"} Check that this diagram commutes and that the new row is exact. ::: Taking the LES for $\Hom(\wait, B)$ yields \begin{tikzcd} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\beta} && \textcolor{rgb,255:red,92;green,92;blue,214}{x} \\ \cdots && {\Hom(P, B)} && {\Hom(P, B)} && {\Ext^1(A, B)} \\ \\ \cdots && {\Hom(X, B)} && {\Hom(B, B)} && {\Ext^1(A, B)} \\ &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\one_B} && \textcolor{rgb,255:red,92;green,92;blue,214}{\Theta(\xi)} \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow[from=4-1, to=4-3] \arrow[from=4-3, to=4-5] \arrow[from=2-5, to=2-7] \arrow[from=4-5, to=4-7] \arrow[no head, from=2-7, to=4-7] \arrow["{\beta_*}"{description}, from=4-5, to=2-5] \arrow["{\sigma_*}"', from=4-3, to=2-3] \arrow["\bd", color={rgb,255:red,92;green,92;blue,214}, maps to, from=5-5, to=5-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, maps to, from=1-5, to=5-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=1-5, to=1-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, dashed, maps to, from=1-7, to=5-7] \end{tikzcd} > $(*)$ [Link to Diagram](https://q.uiver.app/?q=WzAsMTIsWzIsMSwiXFxIb20oUCwgQikiXSxbNCwxLCJcXEhvbShQLCBCKSJdLFswLDEsIlxcY2RvdHMiXSxbNiwxLCJcXEV4dF4xKEEsIEIpIl0sWzIsMywiXFxIb20oWCwgQikiXSxbNCwzLCJcXEhvbShCLCBCKSJdLFs2LDMsIlxcRXh0XjEoQSwgQikiXSxbMCwzLCJcXGNkb3RzIl0sWzQsMCwiXFxiZXRhIixbMjQwLDYwLDYwLDFdXSxbNiwwLCJ4IixbMjQwLDYwLDYwLDFdXSxbNCw0LCJcXG9uZV9CIixbMjQwLDYwLDYwLDFdXSxbNiw0LCJcXFRoZXRhKFxceGkpIixbMjQwLDYwLDYwLDFdXSxbMiwwXSxbMCwxXSxbNyw0XSxbNCw1XSxbMSwzXSxbNSw2XSxbMyw2LCIiLDEseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzUsMSwiXFxiZXRhXyoiLDFdLFs0LDAsIlxcc2lnbWFfKiIsMl0sWzEwLDExLCJcXGJkIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX0sWzI0MCw2MCw2MCwxXV0sWzgsMTAsIiIsMSx7ImN1cnZlIjotNSwiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dLFs4LDksIiIsMSx7ImNvbG91ciI6WzI0MCw2MCw2MF0sInN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XSxbOSwxMSwiIiwxLHsiY3VydmUiOi01LCJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) So we - Started with $x$ - Took a reference SES - Produce the cokernel - Took a pushout and found $\beta$. - Showed that $\beta\mapsto x$. \todo[inline]{Review video: 9:28 AM!} This shows surjectivity, but depended on choice of $\beta$. :::{.claim} $\Theta$ is injective. ::: Note that the previous construction there is a way to associate to $x\in \Ext^1(A, B)$ an extension of $A$ by $B$. To see that this gives a well-defined map $\Psi$, so $\Psi(x) = [ \xi ]$ as well, suppose $\beta'\in \Hom(M, B)$ is another lift of $x$. Note that although $\Ext^1(P, B) =0$, the fact that $\ker \bd = \Hom(M, B) \neq 0$, there are many such choices of lifts. Using exactness of diagram $(*)$, there exists an $f\in \Hom(P, B)$ such that $\beta' = \beta + fj$, recalling that $j: M\to P$. Now taking the pushout $X'$ of $j$ and $\beta'$, the maps $i: B\to X$ and \( \sigma + if: P\to X \) induce an isomorphism $X' \mapsvia{\sim} X$ and thus an equivalence $\xi \mapsvia{\sim} \xi'$. :::{.exercise title="?"} Check this isomorphism. ::: Moreover, given any extension $\xi$, we can fit it into a diagram of the following form: \begin{tikzcd} && 0 & M & P & A & 0 \\ \\ {\xi:} && 0 & B & X & A & 0 \arrow[equals, from=1-6, to=3-6] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \arrow[from=1-6, to=1-7] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=3-5, to=3-6] \arrow[from=3-6, to=3-7] \arrow["{\exists \beta}"{description}, dashed, from=1-4, to=3-4] \arrow["{\exists \sigma}"{description}, dashed, from=1-5, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTEsWzIsMCwiMCJdLFszLDAsIk0iXSxbNCwwLCJQIl0sWzUsMCwiQSJdLFs2LDAsIjAiXSxbMiwyLCIwIl0sWzYsMiwiMCJdLFszLDIsIkIiXSxbNCwyLCJYIl0sWzUsMiwiQSJdLFswLDIsIlxceGk6Il0sWzMsOSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs1LDddLFs3LDhdLFs4LDldLFs5LDZdLFsxLDcsIlxcZXhpc3RzIFxcYmV0YSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsyLDgsIlxcZXhpc3RzIFxcc2lnbWEiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) First we use projectivity of $P$ to get \( \sigma: P\to X \). Then restricting \( \sigma \) to the kernels of \( \pi, \mu \) respectively makes $\beta: M\to B$, so this diagram commutes :::{.exercise title="?"} Check that $X$ is the pushout of $j$ and $\beta$. ::: It follows that $\Psi (\Theta(\xi)) = \xi$ and thus $\Theta$ is injective, making it a bijection. ::: :::{.remark} Note the importance of the reversed directions after taking the Hom! ::: :::{.remark} How can we upgrade this to a group homomorphism? One way is to pull back the group structure from the right-hand side to the left-hand side, but it turns out that Baer worked out an intrinsic group structure around 1934. We can construct the "smallest" extension such that $A$ is a quotient and $B$ is a submodule. ::: :::{.definition title="Baer Sum (1934)"} Suppose we have two extensions of $A$ by $B$: \[ \xi: & 0\to B \mapsvia{i} X \mapsvia{\pi} A \to 0 \\ \xi': & 0\to B \mapsvia{i'} X' \mapsvia{\pi'} A \to 0 \\ .\] Let $X''$ be the **pullback** of $\pi, \pi'$, defined by \[ X'' \da \ts{ (x, x') \in X \cross X' \st \pi(x) = \pi'(x') \in A } ,\] which identifies the two copies of $A$. This fits into a cartesian square \begin{tikzcd} {X''} && {X'} \\ \\ X && A \arrow["{\pi_1}"', from=1-1, to=3-1] \arrow["{\pi_2}", from=1-1, to=1-3] \arrow["{\pi'}"', from=1-3, to=3-3] \arrow["\pi", from=3-1, to=3-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJYJyciXSxbMiwwLCJYJyJdLFsyLDIsIkEiXSxbMCwyLCJYIl0sWzAsMywiXFxwaV8xIiwyXSxbMCwxLCJcXHBpXzIiXSxbMSwyLCJcXHBpJyIsMl0sWzMsMiwiXFxwaSJdLFswLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) Note that $X''$ contains 3 copies of $B$: - $B \cross 0$, or really $i(B) \cross \ts{ 0 } \subset X''$ (using exactness). - $0\cross B$, i.e. \( \ts{ 0 } \cross i'(B) \subseteq X'' \) (using exactness). - \( \tilde\diagonal = \ts{ (-b, b) \st b\in B } \), the **skew diagonal**. One can check that $\pi i (-b) = 0 = \pi' i' (b)$. Note that we're identifying $B$ with $i(B), i'(B)$. Set $Y \da X'' / \tilde\diagonal$, then $(b, 0) + (-b, b) = (0, b)$ where $(-b, b) \in \tilde \diagonal$, so $B \cross 0$ and \( 0 \cross B \) have the same image in $Y$, since \[ (B \cross 0) \intersect\tilde\diagonal = \ts{ (0, 0) } = (0 \cross B) \intersect\tilde\diagonal .\] In fact this image in $Y$ is isomorphic to $B$, by construction of what we're quotienting out by. Denoting this subgroup of $Y$ by $B$, we get a SES \[ \phi: 0\to B \to Y \to Y/B \to 0 .\] What is $Y/B$? We can write this as \[ Y/B = { X'' / \tilde \diagonal \over (0 \cross B ) / \tilde\diagonal } \cong {X'' \over (0 \cross B) + \tilde\diagonal} \cong {X'' / 0 \cross B \over (\tilde\diagonal + (0 \cross B) ) / (0 \cross B)} .\] But the numerator is isomorphic to $X$ by $\pi_1$, and the denominator is isomorphic to $B$ by $\pi_1$. So $\phi$ is an extension of $A$ by $B$ called the **Baer sum** of $\xi, \xi'$. ::: :::{.corollary title="?"} The equivalence classes of extensions of $A$ by $B$ is an abelian group under Baer sums, where zero is the class of split extensions. Moreover, the map $\Theta$ from the previous theorem is an isomorphism of abelian groups. ::: :::{.remark} Next time we'll check this by showing $\Theta(\phi) = \Theta(\xi) + \Theta(\xi')$. :::