----- > [!proposition] Proposition. ([[five lemma]]) > Consider the following commutative [[diagram]] in $R$-$\mathsf{Mod}$.[^1] > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAEEQBfU9TXfIRQBGclVqMWbAELdeIDNjwEiAJjHV6zVohABhOXyWCiAZg0TtbACKGF-ZUOQAWC1qm6AoncUCVKMmFxdx0OAHIfBxMRUiDNSVDpCJ4jPyd1OMsPfWT5X0czWOCEm1zUgpRXTJC2T2TxGCgAc3giUAAzACcIAFskMhAcCCRRLND2uy7ekeohpHUxtibJ7r7EBbnEc0XdAAsV6a3Z4cRXHZAAKwO1gFZjpAA2eKtdTpBqBjoAIxgGAAUov4QJ0sE1djhro97ogAOzPbIID7fX4A4xAkFgiEpEBTNZwwYnAAc8NCEKRP3+gKEwNB4MhiGJBKQAE4SWwmO8QJ8Kai0mwMXTsbj+tC7ucADri34MTnclFUtgMGDtMkgXYwOhQNg4ADuEHVmoQQtWMyZiCe5x6pAABJLsH1jYcNid8TVdIRHWttptGW6FDa7VgHfJhadoazzgBHTk4OhYGV7CAQADW3AoXCAA > \begin{tikzcd} > A \arrow[r, "f"] \arrow[d, "\ell", two heads] & B \arrow[r, "g"] \arrow[d, "{m, \sim}"] & C \arrow[r, "h"] \arrow[d, "n"] & D \arrow[r, "j"] \arrow[d, "{p, \sim}"] & E \arrow[d, "q", hook] \\ > A' \arrow[r, "r"'] & B' \arrow[r, "s"'] & C' \arrow[r, "t"'] & D' \arrow[r, "u"'] & E' > \end{tikzcd} > \end{document} > ``` > where we have assumed the two rows are [[exact sequence|exact]], $m$ and $p$ are [[isomorphism|isomorphisms]], $\ell$ is an [[epimorphism]] and $q$ is a [[monomorphism]]. > > Then $n$ is also an [[isomorphism]]. > > The five lemma (recalling epic + monic $\implies$ iso. in $R$-$\mathsf{Mod}$) follows by combining the two (dual) *four lemmas*: > > **1.** Given a [[diagram]]: > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZARgBoAGAXVJADcBDAGwFcYkQAhEAX1PU1z5CKAEwVqdJq3YBhHnxAZseAkQDM4mgxZtEIACLz+yoUQAsmyTvYBRI4oErhJUsQnbpejgHJ7Swaqiru5SuiAyvrzGAc4ablqh7PqRCv5O5sEJ1no2KdHpKOSWHmEAOqV+jqaFmVaeIOU8EjBQAObwRKAAZgBOEAC2SEUgOBBIZHVhrfa9A+M0o0hik+wAFjN9g4jLi4gaK3oAVhtziBYjY4gArFn1CDSM9ABGMIwAClWBID1Yras4Jy2NwuSAAbLcwgCHs9Xh8TF8fn8AVEQLMtuCQYgAOwQ9jMEDQl7vT7Cb6-f6AoYLS7nErsQY0VYwehQdg4ADuECZLIQKLR80xwLpekIfM2S2pYNxejQBJA3NZeg5XOZUF5Cn5e0l2OlIAAjnKcPQsIw1hAIABrJrcIA > \begin{tikzcd} > \ & B \arrow[r, "g"] \arrow[d, "m", two heads] & C \arrow[r, "h"] \arrow[d, "n"] & D \arrow[r, "j"] \arrow[d, "p", two heads] & E \arrow[d, "q", hook] \\ > \ & B' \arrow[r, "s"'] & C' \arrow[r, "t"'] & D' \arrow[r, "u"'] & E' > \end{tikzcd} > \end{document} > ``` > $n$ is an [[epimorphism]]. > > **2.** Given a [[diagram]]: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAEEQBfU9TXfIRQBGclVqMWbAELdeIDNjwEiAJjHV6zVohABhOXyWCiAZg0TtbACKGF-ZUJKlh4rVN3sA5HcUCVIi5ukjog0j48Rv5O6q6aIWx6EfJ+jmZB8Va61slRaSgALBbuoQA6pb4OJoUZlh4g5dziMFAA5vBEoABmAE4QALZIZCA4EEiidaFddr0D49SjSOqTbK0zfYOIy4uI5iu6ABbrc4hFI2OIAKyZ9T0g1Ax0AEYwDAAKVQEgPVitBzjHTbXc5IABsN1CCAez1eH2MXx+fwBkRAs024JBiAA7BC2ADoS93p8hN9fv9AUMFhcziU2OVXgx7iBHoS4dE2AwYF18SADjA6FA8QB3CB8gUIFFo+aY4G03SDBZ0LCMw4QCAAawpWypYNxukIko2SD2Oxx+wUTJwSpVvLVmq4FC4QA > \begin{tikzcd} > A \arrow[r, "f"] \arrow[d, "\ell", two heads] & B \arrow[r, "g"] \arrow[d, "m", hook] & C \arrow[r, "h"] \arrow[d, "n"] & D \arrow[d, "p", hook] & \ \\ > A' \arrow[r, "r"'] & B' \arrow[r, "s"'] & C' \arrow[r, "t"'] & D' & \ > \end{tikzcd} > \end{document} > ``` > $n$ is a [[monomorphism]]. > [!proof]- Proof. ([[five lemma]]) > Assume we are in the [[category]] $R$-$\mathsf{Mod}$. Note this means $\ell$ [[module homomorphism is surjective iff cokernel is trivial iff is an epimorphism|is a]] [[surjection]] and $q$ [[module homomorphism is injective iff kernel is trivial iff is a monomorphism|is an]] [[injection]]. (Later justify why this is enough for any [[abelian category]] and $\mathsf{Grp}$.) > > **Proof of the 1st four lemma.** Assume $m$ and $p$ are [[surjection|surjective]] and $q$ is [[injection|injective]]. Let $c' \in C'$. Want to find an element of $C$ that maps to $c'$ under $n$. Proceed to diagram chase (draw things out to follow along): > 1. Since $p$ is surjective, there exists $d \in D$ with $p(d)=t(c')$. > 2. Since $\operatorname{im }t=\operatorname{ker }u$ by exactness, $u(p(d))=u(t(c'))=0$. > 3. By commutativity of the diagram, $u(p(d))=q(j(d))$. Hence $q(j(d))=0$, i.e., $j(d) \in \operatorname{ker }q$. > 4. $q$ is injective, so $j(d)=0$, i.e., $d \in \operatorname{ker }j=\operatorname{im }h$. Fix $c \in C$ with $h(c)=d$. > 5. By commutativity of the diagram, $t(n(c))=p(h(c))$. > 6. Since $p(h(c))=p(d)=t(c')$, this means $t(n(c))=t(c')$. Rearranging: $t\big( c' - n (c) \big)=0$. > 7. Thus, $c'-n(c) \in \operatorname{ker }t=\operatorname{im }s$. Fix $b' \in B'$ with $s(b')=c'-n(c)$. > 8. $m$ is surjective, so there exists $b \in B$ with $m(b)=b'$. > 9. By commutativity of the diagram, $s(m(b))=n(g(b))$. > 10. Combining $(8)$ and $(9)$ gives us $s(b')=s(m(b))=n(g(b))$. By $(7)$, $s(b')=c'-n(c)$. Thus, $c'-n(c)=n(g(b))$. > 11. Rearranging: $c'=n\big( g(b) +c \big)$. Thus, $n$ is surjective. > > > > > **** > > **Proof of the 2nd four lemma.** > Exercise. ----- #### [^1]: The result holds in any [[abelian category]], and also [[group|in]] $\mathsf{Grp}$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```