From bb135dc4be57e7fb4a628bb240361f96adfa5a4c Mon Sep 17 00:00:00 2001 From: zeramorphic Date: Tue, 17 Oct 2023 10:52:13 +0100 Subject: [PATCH] Some more proofs Signed-off-by: zeramorphic --- iii/cat/02_yoneda_lemma.tex | 2 +- iii/commalg/02_tensor_products.tex | 45 +++++++++++++++++++++++++++--- 2 files changed, 42 insertions(+), 5 deletions(-) diff --git a/iii/cat/02_yoneda_lemma.tex b/iii/cat/02_yoneda_lemma.tex index 1ed0341..d6cafa2 100644 --- a/iii/cat/02_yoneda_lemma.tex +++ b/iii/cat/02_yoneda_lemma.tex @@ -292,7 +292,7 @@ \subsection{Separating and detecting families} A & B & D \arrow["\ell", shift left=2, from=2-2, to=2-3] \arrow["m"', shift right=2, from=2-2, to=2-3] - \arrow["f", from=2-1, to=2-2] + \arrow["f"', from=2-1, to=2-2] \arrow["n", from=1-2, to=2-2] \arrow[dashed, from=1-2, to=2-1] \end{tikzcd}\] diff --git a/iii/commalg/02_tensor_products.tex b/iii/commalg/02_tensor_products.tex index 9859d22..5c12ab4 100644 --- a/iii/commalg/02_tensor_products.tex +++ b/iii/commalg/02_tensor_products.tex @@ -718,7 +718,6 @@ \subsection{Exactness properties of the tensor product} from the category of \( R \)-modules to itself given by \[ T_M(N) = M \otimes_R N;\quad T_M(N \xrightarrow f N') = \id_M \otimes f \] We intend to show that if -% https://q.uiver.app/#q=WzAsNCxbMCwwLCJBIl0sWzEsMCwiQiJdLFsyLDAsIkMiXSxbMywwLCIwIl0sWzAsMSwiZiJdLFsxLDIsImciXSxbMiwzXV0= \[\begin{tikzcd} A & B & C & 0 \arrow["f", from=1-1, to=1-2] @@ -726,7 +725,6 @@ \subsection{Exactness properties of the tensor product} \arrow[from=1-3, to=1-4] \end{tikzcd}\] is an exact sequence of \( R \)-modules, then -% https://q.uiver.app/#q=WzAsNCxbMCwwLCJNIFxcb3RpbWVzX1IgQSJdLFsxLDAsIk0gXFxvdGltZXNfUiBCIl0sWzIsMCwiTSBcXG90aW1lc19SIEMiXSxbMywwLCIwIl0sWzAsMSwiXFxpZF9NIFxcb3RpbWVzIGYiXSxbMSwyLCJcXGlkX00gXFxvdGltZXMgZyJdLFsyLDNdXQ== \[\begin{tikzcd} {M \otimes_R A} & {M \otimes_R B} & {M \otimes_R C} & 0 \arrow["{T_M(f)}", from=1-1, to=1-2] @@ -744,7 +742,11 @@ \subsection{Exactness properties of the tensor product} \end{definition} \begin{definition} Let \( Q, P \) be \( R \)-modules. - Then \( \Hom_R(Q, -) \) and \( \Hom_R(-, P) \) are functors \( \mathbf{Mod}_R \to \mathbf{Mod}_R \), with action on morphisms \( f : N' \to N \) given by + Then + \[ \Hom_R(Q, -) : \mathbf{Mod}_R \to \mathbf{Mod}_R \] + and + \[ \Hom_R(-, P) : \mathbf{Mod}_R^\cop \to \mathbf{Mod}_R \] + are functors, with action on morphisms \( f : N' \to N \) given by \[ \Hom_R(Q, f)(\varphi) = f \circ \varphi = f_\star(\varphi) : \Hom_R(Q, N') \to \Hom_R(Q, N') \] and \[ \Hom_R(f, P)(\varphi) = \varphi \circ f = f^\star(\varphi) : \Hom_R(N, Q) \to \Hom_R(N', Q) \] @@ -767,6 +769,22 @@ \subsection{Exactness properties of the tensor product} \end{tikzcd}\] Thus, the covariant hom-functor is \emph{left exact}. \end{proposition} +\begin{proof} + First, we show \( f_\star \) is injective. + Suppose \( f_\star(\varphi) = 0 \), so \( f \circ \varphi = 0 \). + Then as \( f \) is injective, \( f(\varphi(x)) = 0 \) implies \( \varphi(x) = 0 \), giving \( \varphi = 0 \) as required. + + Now consider \( \varphi : Q \to A \). + Then + \[ g_\star (f_\star(\varphi)) = g \circ (f \circ \varphi) = (g \circ f) \circ \varphi = 0 \circ \varphi = 0 \] + so \( \Im f_\star \subseteq \ker g_\star \). + Now suppose \( \varphi : Q \to B \) has \( g_\star(\varphi) = g \circ \varphi = 0 \). + So for all \( x \in Q \), \( g(\varphi(x)) = 0 \). + By exactness of the original sequence, \( \varphi(x) \in \Im f \). + As \( f \) is injective, \( \varphi(x) \) has a unique preimage \( \psi(x) \) under \( f \). + As \( f \) is \( R \)-linear, so is \( \psi : Q \to A \). + Hence \( f_\star(\psi) = \varphi \) as required. +\end{proof} \begin{proposition} Suppose \[\begin{tikzcd} @@ -785,6 +803,25 @@ \subsection{Exactness properties of the tensor product} \end{tikzcd}\] Thus, the contravariant hom-functor is also left-exact. \end{proposition} +\begin{proof} + First, we show \( g^\star \) is injective. + Suppose \( g^\star(\varphi) = 0 \), so \( \varphi \circ g = 0 \). + As \( g \) is surjective, we must have \( \varphi = 0 \). + + Now consider \( \varphi : C \to P \). + Then + \[ f^\star(g^\star(\varphi)) = (\varphi \circ g) \circ f = \varphi \circ (g \circ f) = \varphi \circ 0 = 0 \] + so \( \Im g^\star \subseteq \ker f^\star \). + Now suppose \( \varphi : B \to P \) has \( f^\star(\varphi) = \varphi \circ f = 0 \). + So for all \( x \in A \), \( \varphi(f(x)) = 0 \). + Define \( \psi : C \to P \) by + \[ \psi(g(x)) = \varphi(x) \] + We show this is well-defined. + If \( g(x) = g(y) \), then \( g(x - y) = 0 \), so \( x - y = f(a) \) for some \( a \in A \). + But then \( \varphi(f(a)) = 0 \), so \( \varphi(x) = \varphi(y) \). + As \( \varphi \) and \( g \) are ring homomorphisms, so is \( \psi \). + Hence \( g^\star(\psi) = \varphi \) as required. +\end{proof} \begin{lemma} Consider a sequence of \( R \)-modules \[\begin{tikzcd} @@ -819,7 +856,7 @@ \subsection{Exactness properties of the tensor product} \[ \id_C \mapsto \id_C \circ g \mapsto \id_C \circ g \circ f \] By exactness, \( \id_C \) must be mapped to zero under \( f^\star \circ g^\star \), so \( g \circ f = 0 \). Hence \( \Im f \subseteq \ker g \). - + Now, take \( P = \faktor{B}{\Im f} = \coker f \). \[\begin{tikzcd} {\Hom_R\qty(C, \faktor{B}{\Im f})} & {\Hom_R\qty(B, \faktor{B}{\Im f})} & {\Hom_R\qty(A, \faktor{B}{\Im f})}