Some more proofs

Signed-off-by: zeramorphic <[email protected]>

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}\]

iii/commalg/02_tensor_products.tex

@@ -718,15 +718,13 @@ \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=
 \[\begin{tikzcd}
 	A & B & C & 0
 	\arrow["f", from=1-1, to=1-2]
 	\arrow["g", from=1-2, to=1-3]
 	\arrow[from=1-3, to=1-4]
 \end{tikzcd}\]
 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})}