Lowercase im

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

iii/alggeom/02_sheaves.tex

@@ -173,11 +173,11 @@ \subsection{Sheafification}
 
 \subsection{Kernels and cokernels}
 Let \( \varphi : \mathcal F \to \mathcal G \) be a morphism of presheaves.
-Then we can define presheaves \( \ker \varphi, \coker \varphi, \Im \varphi \) by
+Then we can define presheaves \( \ker \varphi, \coker \varphi, \im \varphi \) by
 \begin{align*}
     (\ker \varphi)(U) &= \ker \varphi_U \\
     (\coker \varphi)(U) &= \coker \varphi_U \\
-    (\Im \varphi)(U) &= \Im \varphi_U
+    (\im \varphi)(U) &= \im \varphi_U
 \end{align*}
 One can check that these are indeed presheaves.
 \begin{proposition}

iii/alggeom/06_divisors.tex

@@ -125,7 +125,7 @@ \subsection{Cartier divisors}
 \begin{definition}
     The image of \( \Gamma(X, \mathcal K_X^\star) \) in \( \Gamma\qty(X, \faktor{\mathcal K_X^\star}{\mathcal O_X^\star}) \) is the set of \emph{principal} Cartier divisors.
     The \emph{Cartier class group} is the quotient
-    \[ \faktor{\Gamma\qty(X, \faktor{\mathcal K_X^\star}{\mathcal O_X^\star})}{\Im \Gamma(X, \mathcal K_X^\star)} \]
+    \[ \faktor{\Gamma\qty(X, \faktor{\mathcal K_X^\star}{\mathcal O_X^\star})}{\im \Gamma(X, \mathcal K_X^\star)} \]
 \end{definition}
 A section \( \mathcal D \in \Gamma\qty(X, \faktor{\mathcal K_X^\star}{\mathcal O_X^\star}) \) can be specified by \( \qty{(U_i, f_i)} \) where the \( \qty{U_i} \) form an open cover and \( f_i \in \mathcal K_X^\star(U_i) \), such that on \( U_i \cap U_j \), the quotient \( \frac{f_i}{f_j} \) lies in \( \mathcal O_X^\star(U_i \cap U_j) \).
 

iii/alggeom/07_sheaf_cohomology.tex

@@ -52,7 +52,7 @@ \subsection{\v{C}ech cohomology}
 Thus, \( \qty{C^p(\mathcal U, \mathcal F)}_p \) has the structure of a \emph{cochain complex}.
 \begin{definition}
 	The \emph{\( i \)th \v{C}ech cohomology} of \( (X, \mathcal F, \mathcal U) \) is the \( i \)th cohomology group of the cochain complex:
-	\[ \check{H}^i(X, \mathcal F) = \frac{\ker(C^i(\mathcal U, \mathcal F) \xrightarrow d C^{i+1}(\mathcal U, \mathcal F))}{\Im(C^{i-1}(\mathcal U, \mathcal F) \xrightarrow d C^i(\mathcal U, \mathcal F))} \]
+	\[ \check{H}^i(X, \mathcal F) = \frac{\ker(C^i(\mathcal U, \mathcal F) \xrightarrow d C^{i+1}(\mathcal U, \mathcal F))}{\im(C^{i-1}(\mathcal U, \mathcal F) \xrightarrow d C^i(\mathcal U, \mathcal F))} \]
 \end{definition}
 \begin{example}
 	Let \( X = S^1 \) be the usual circle.
@@ -112,7 +112,7 @@ \subsection{\v{C}ech cohomology}
 	Thus,
 	\begin{align*}
 		H^n(\mathbb P^n_k, \mathcal F) &= \check{H}^n(\mathcal U, \mathcal F) \\
-		&= \frac{\check{C}^n}{\Im (\check{C}^{n-1} \to \check{C}^n)} \\
+		&= \frac{\check{C}^n}{\im (\check{C}^{n-1} \to \check{C}^n)} \\
 		&= \frac{\vecspan_k \qty{x_0^{k_0} \dots x_n^{k_n} \mid k_i \in \mathbb Z}}{\vecspan_k \qty{x_0^{k_0} \dots x_n^{k_n} \mid \text{at least one } k_i \geq 0}}
 	\end{align*}
 	as required.

iii/cat/07_additive_and_abelian_categories.tex

@@ -191,14 +191,14 @@ \subsection{Abelian categories}
         \item The category \( \mathbf{AbGp} \) is abelian; more generally, for any ring \( R \), the category \( \mathbf{Mod}_R \) is abelian.
         \item If \( \mathcal A \) is abelian and \( \mathcal C \) is small, then \( [\mathcal C, \mathcal A] \) is abelian, with all structures defined pointwise.
         \item If \( \mathcal A \) is abelian and \( \mathcal C \) is small and additive, then the category of additive functors \( \mathcal C \to \mathcal A \), denoted \( \operatorname{Add}(\mathcal C, \mathcal A) \), is also abelian, as it is closed under all of the structures on \( [\mathcal C, \mathcal A] \).
-        Note that this covers the case of \( R \)-modules, as an additive category with a single object is a ring, and the category of modules over such a ring is isomorphic to the category of additive functors from this category to \( \mathbf{AbGp} \)
+        Note that this covers the case of \( R \)-modules, as an additive category with a single object is a ring, and the category of modules over such a ring is isomorphic to the category of additive functors from this category to \( \mathbf{AbGp} \).
     \end{enumerate}
 \end{example}
 \begin{remark}
     If \( f : A \to B \) in an abelian category, then \( \ker \coker f \) is the smallest subobject \( I \rightarrowtail B \) through which \( f \) factors.
-    This is called the \emph{image} of \( f \), denoted \( \Im f = \ker \coker f \).
+    This is called the \emph{image} of \( f \), denoted \( \im f = \ker \coker f \).
     The other part of the factorisation \( A \to I \) is epic, as it cannot factor through the equaliser of any nonequal parallel pair \( I \rightrightarrows C \).
-    Thus, it is also the smallest quotient of \( A \) through which \( f \) factors, so it is the \emph{coimage} of \( f \), given by \( \coker \ker f \).
+    Thus, it is also the smallest quotient of \( A \) through which \( f \) factors, so it is the \emph{coimage} of \( f \), given by \( \coim f = \coker \ker f \).
     The composition \( A \twoheadrightarrow I \rightarrowtail B \) is the unique epi--mono factorisation of \( f \).
 \end{remark}
 To show that this factorisation is stable under pullback, it suffices to show that the pullback of an epimorphism in an abelian category is epic, as the corresponding statement for monomorphisms has already been shown.
@@ -335,7 +335,7 @@ \subsection{Exact sequences}
     in an abelian category \( \mathcal A \) is \emph{exact} at \( A_n \) if \( \ker f_n = \operatorname{im} f_{n+1} \).
     The entire sequence is said to be \emph{exact} if it is exact at every vertex.
 \end{definition}
-By duality, the sequence is exact at \( A_n \) if and only if \( \coker f_{n+1} = \operatorname{coim} f_n \).
+By duality, the sequence is exact at \( A_n \) if and only if \( \coker f_{n+1} = \coim f_n \).
 \begin{example}
     % https://q.uiver.app/#q=WzAsNCxbMCwwLCIwIl0sWzEsMCwiQSJdLFsyLDAsIkIiXSxbMywwLCJDIl0sWzAsMV0sWzEsMiwiZiJdLFsyLDMsImciXV0=
 \[\begin{tikzcd}
@@ -476,7 +476,7 @@ \subsection{The five lemma}
 	\arrow["{f_4}", from=2-4, to=2-5]
 	\arrow["{u_5}", from=2-5, to=3-5]
 \end{tikzcd}\]
-    Then \( y \) is also the pullback of this factorisation of \( x \) along \( \operatorname{coim} f_2 \), so \( y \) is an epimorphism as epimorphisms are stable under pullback.
+    Then \( y \) is also the pullback of this factorisation of \( x \) along \( \coim f_2 \), so \( y \) is an epimorphism as epimorphisms are stable under pullback.
     Then \( g_2 u_2 z = u_3 f_2 z = u_3 x y = 0 \).
     Thus \( u_2 z \) factors through \( \ker g_2 = \operatorname{im} g_1 \).
     Consider the pullback square
@@ -489,7 +489,7 @@ \subsection{The five lemma}
 	\arrow["w"', from=1-1, to=2-1]
 	\arrow["{g_1 u_1}"', from=2-1, to=2-2]
 \end{tikzcd}\]
-    So \( v \) is epic, as it is the pullback of \( \operatorname{coim}(g_1 u_1) \).
+    So \( v \) is epic, as it is the pullback of \( \coim(g_1 u_1) \).
     % https://q.uiver.app/#q=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
 \[\begin{tikzcd}
 	E & D & C \\
@@ -624,7 +624,7 @@ \subsection{Complexes in abelian categories}
 \end{tikzcd}\]
     By definition, \( I_n \to C_n \) is \( \ker (C_n \to Q_n) \).
     As \( Z_n \to C_n \) is a monomorphism, \( I_n \to Z_n \) is \( \ker (Z_n \to C_n \to Q_n) \).
-    Hence \( Z_n \to H_n \) is \( \operatorname{coim}(Z_n \to Q_n) \), so we obtain
+    Hence \( Z_n \to H_n \) is \( \coim(Z_n \to Q_n) \), so we obtain
     % https://q.uiver.app/#q=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
 \[\begin{tikzcd}
 	{C_{n+1}} && {C_n} && {C_{n-1}} \\

iii/commalg/02_tensor_products.tex

@@ -783,10 +783,10 @@ \subsection{Exactness properties of the tensor product}
     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 \( \Im f_\star \subseteq \ker g_\star \).
+    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, \( \varphi(x) \in \Im f \).
+    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.
@@ -817,7 +817,7 @@ \subsection{Exactness properties of the tensor product}
     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 \( \Im g^\star \subseteq \ker f^\star \).
+    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
@@ -861,19 +861,19 @@ \subsection{Exactness properties of the tensor product}
     Consider
     \[ \id_C \mapsto \id_C \circ g \mapsto \id_C \circ g \circ f \]
     By exactness, \( \id_C \) must be mapped to zero under \( f^\star \circ g^\star \), so \( g \circ f = 0 \).
-    Hence \( \Im f \subseteq \ker g \).
+    Hence \( \im f \subseteq \ker g \).
 
-    Now, take \( P = \faktor{B}{\Im f} = \coker f \).
+    Now, take \( P = \faktor{B}{\im f} = \coker f \).
     \[\begin{tikzcd}
-        {\Hom_R\qty(C, \faktor{B}{\Im f})} & {\Hom_R\qty(B, \faktor{B}{\Im f})} & {\Hom_R\qty(A, \faktor{B}{\Im f})}
+        {\Hom_R\qty(C, \faktor{B}{\im f})} & {\Hom_R\qty(B, \faktor{B}{\im f})} & {\Hom_R\qty(A, \faktor{B}{\im f})}
         \arrow["{g^\star}", from=1-1, to=1-2]
         \arrow["{f^\star}", from=1-2, to=1-3]
     \end{tikzcd}\]
-    Let \( h : B \to \faktor{B}{\Im f} \) be the quotient map.
+    Let \( h : B \to \faktor{B}{\im f} \) be the quotient map.
     Then,
     \[ f^\star(h) = h \circ f;\quad h(f(x)) = 0 \]
-    Thus by exactness, \( h \) has a preimage \( e : C \to \faktor{B}{\Im f} \).
-    Then \( g^\star(e) = e \circ g = h \), so \( \ker g \subseteq \ker h = \Im f \), giving the reverse inclusion.
+    Thus by exactness, \( h \) has a preimage \( e : C \to \faktor{B}{\im f} \).
+    Then \( g^\star(e) = e \circ g = h \), so \( \ker g \subseteq \ker h = \im f \), giving the reverse inclusion.
 \end{proof}
 By the universal property of the tensor product,
 \[ \Hom_R(M \otimes_R N, L) \simeq \operatorname{Bilin}_R(M \times N, L) \simeq \Hom_R(N, \Hom_R(M, L)) \]
@@ -1086,9 +1086,9 @@ \subsection{Flat modules}
     % https://q.uiver.app/#q=WzAsMTMsWzIsMiwiQSJdLFs0LDIsIkIiXSxbNiwyLCJDIl0sWzEsMSwiXFxrZXIgZiJdLFszLDMsIlxcSW0gZiA9IFxca2VyIGciXSxbNCw0LCIwIl0sWzAsMCwiMCJdLFsyLDQsIjAiXSxbNSwxLCJcXEltIGciXSxbNiwwLCIwIl0sWzQsMCwiMCJdLFs3LDMsIlxcY29rZXIgZyJdLFs4LDQsIjAiXSxbMCwxLCJmIl0sWzEsMiwiZyJdLFszLDBdLFswLDRdLFs0LDVdLFs2LDNdLFs3LDRdLFs0LDFdLFsxLDhdLFs4LDldLFsxMCw4XSxbOCwyXSxbMiwxMV0sWzExLDEyXV0=
 \[\begin{tikzcd}[column sep=small]
 	0 &&&& 0 && 0 \\
-	& {\ker f} &&&& {\Im g} \\
+	& {\ker f} &&&& {\im g} \\
 	&& A && B && C \\
-	&&& {\Im f = \ker g} &&&& {\coker g} \\
+	&&& {\im f = \ker g} &&&& {\coker g} \\
 	&& 0 && 0 &&&& 0
 	\arrow["f", from=3-3, to=3-5]
 	\arrow["g", from=3-5, to=3-7]
@@ -1107,10 +1107,10 @@ \subsection{Flat modules}
 \end{tikzcd}\]
     After applying \( T = T_M \), the diagram still commutes, and the diagonal lines remain exact.
     \begin{align*}
-        \Im(TA \to TB) &= \Im(TA \to T(\Im f) \to TB) \\
-        &= \Im (T(\Im f) \to TB) \\
-        &= \ker (TB \to T(\Im g)) \\
-        &= \ker (TB \to T(\Im g) \to TC) \\
+        \im(TA \to TB) &= \im(TA \to T(\im f) \to TB) \\
+        &= \im (T(\im f) \to TB) \\
+        &= \ker (TB \to T(\im g)) \\
+        &= \ker (TB \to T(\im g) \to TC) \\
         &= \ker (TB \to TC)
     \end{align*}
 

iii/commalg/03_localisation.tex

@@ -228,11 +228,11 @@ \subsection{Exactness}
 \begin{proof}
     First,
     \[ (S^{-1}g) \circ (S^{-1}f) = S^{-1}(g \circ f) = S^{-1}0 = 0 \]
-    so \( \Im S^{-1}f \subseteq \ker S^{-1}g \).
+    so \( \im S^{-1}f \subseteq \ker S^{-1}g \).
     Now suppose \( \frac{b}{s} \in \ker S^{-1}g \), so \( \frac{g(b)}{s} = \frac{0}{1} \).
     Hence there exists \( u \in S \) such that \( ug(b) = 0 \).
     As \( g \) is \( R \)-linear and \( u \in R \), we have \( g(ub) = 0 \).
-    By exactness, \( ub \in \ker g = \Im f \).
+    By exactness, \( ub \in \ker g = \im f \).
     Thus there exists \( a \in A \) such that \( f(a) = ub \).
     Hence,
     \[ \frac{b}{s} = \frac{ub}{us} = \frac{f(a)}{us} = S^{-1}f\qty(\frac{a}{us}) \]
@@ -387,7 +387,7 @@ \subsection{Extension and contraction of ideals}
     \[ \varphi : S^{-1}R \to FF\qty(\faktor{R}{\mathfrak p});\quad \frac{r}{s} \mapsto \frac{r + \mathfrak p}{s + \mathfrak p} \]
     It suffices to show that \( \ker \varphi = \mathfrak p^e \), then the result holds by the isomorphism theorem.
     Let \( \frac{r}{s} \in \ker \varphi \), so \( \frac{r + \mathfrak p}{s + \mathfrak p} = \frac{0}{1} \) in \( FF\qty(\faktor{R}{\mathfrak p}) \).
-    Observe that \( \Im \varphi \subseteq \overline S^{-1} \qty(\faktor{R}{\mathfrak p}) \), where \( \overline S \) is the image of \( S \) in \( \faktor{R}{\mathfrak p} \).
+    Observe that \( \im \varphi \subseteq \overline S^{-1} \qty(\faktor{R}{\mathfrak p}) \), where \( \overline S \) is the image of \( S \) in \( \faktor{R}{\mathfrak p} \).
     Restricting the range, we can consider \( \varphi \) as a map from \( S^{-1}R \) to \( \overline S^{-1}\qty(\faktor{R}{\mathfrak p}) \).
     So \( \varphi\qty(\frac{r}{s}) = \frac{0}{1} \) implies that there exists \( u + \mathfrak p \in \overline S \) such that \( (u + \mathfrak p)(r + \mathfrak p) = 0 \), so \( ur + \mathfrak p = 0 \).
     In particular, \( u \in S \) and \( ur \in \mathfrak p \).

iii/mtncl/02_quantifier_elimination.tex

@@ -199,7 +199,7 @@ \subsection{Amalgamation}
     \[ \mathcal T' = \operatorname{Diag}_{\text{el}} \mathcal M \cup \bigcup_{s \in S} \qty{s = f(s)} \]
     Taking the conjunction, we can suppose it is a single formula \( \varphi(\vb n) \), where \( \vb n \in \mathcal N \) is a tuple of pairwise distinct elements.
     \[ \mathcal T' \vdash \neg\varphi(\vb n) \]
-    Then, using the sentences \( s = f(s) \) and the fact that \( \langle S \rangle_{\mathcal M} \) is generated by \( S \), the formula \( \varphi(\vb n) \) is equivalent modulo \( \mathcal T' \) to some quantifier-free formula \( \psi(\vb s, \vb n') \) where \( \vb s \in S \) and \( \vb n' \in \mathcal N \setminus \Im f \).
+    Then, using the sentences \( s = f(s) \) and the fact that \( \langle S \rangle_{\mathcal M} \) is generated by \( S \), the formula \( \varphi(\vb n) \) is equivalent modulo \( \mathcal T' \) to some quantifier-free formula \( \psi(\vb s, \vb n') \) where \( \vb s \in S \) and \( \vb n' \in \mathcal N \setminus \im f \).
     \[ \mathcal T' \vdash \neg\psi(\vb s, \vb n') \]
     Now, note that \( \mathcal T' \) has nothing to say about \( \vb n' \), so in fact
     \[ \mathcal T' \vdash \forall \vb x.\, \neg\psi(\vb s, \vb x) \]

util.sty

@@ -210,7 +210,9 @@
 \DeclareMathOperator{\dom}{dom}
 \DeclareMathOperator{\fchar}{char}
 \DeclareMathOperator{\code}{code}
+\DeclareMathOperator{\im}{im}
 \DeclareMathOperator{\coker}{coker}
+\DeclareMathOperator{\coim}{coim}
 \DeclareMathOperator{\End}{End}
 \DeclareMathOperator{\ob}{ob}
 \DeclareMathOperator{\mor}{mor}