Replace ◦ with \circ in math mode

src/content/1.1/Category - The Essence of Composition.tex

@@ -34,7 +34,7 @@ \section{Arrows as Functions}\label{arrows-as-functions}
 \end{Verbatim}
 or the chevron \code{>>} in F\#, which both
 go from left to right. But in mathematics and in Haskell functions
-compose right to left. It helps if you read \code{g◦f} as ``g \emph{after} f.''
+compose right to left. It helps if you read \code{g \ensuremath{\circ} f} as ``g \emph{after} f.''
 
 Let's make this even more explicit by writing some C code. We have one
 function \code{f} that takes an argument of type \code{A} and
@@ -83,7 +83,7 @@ \section{Arrows as Functions}\label{arrows-as-functions}
 composition as:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-g ◦ f
+g \ensuremath{\circ} f
 \end{Verbatim}
 You can even use Unicode double colons and arrows:
 
@@ -108,7 +108,7 @@ \section{Properties of Composition}\label{properties-of-composition}
 expressed as:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-h◦(g◦f) = (h◦g)◦f = h◦g◦f
+h \ensuremath{\circ} (g \ensuremath{\circ} f) = (h \ensuremath{\circ} g) \ensuremath{\circ} f = h \ensuremath{\circ} g \ensuremath{\circ} f
 \end{Verbatim}
 In (pseudo) Haskell:
 
@@ -131,11 +131,11 @@ \section{Properties of Composition}\label{properties-of-composition}
 object A is called \code{id\textsubscript{A}} (\newterm{identity} on A). In math
 notation, if \code{f} goes from A to B then
 
-\code{f◦id\textsubscript{A} = f}
+\code{f \ensuremath{\circ} id\textsubscript{A} = f}
 
 and
 
-\code{id\textsubscript{B}◦f = f}
+\code{id\textsubscript{B} \ensuremath{\circ} f = f}
 \end{enumerate}
 When dealing with functions, the identity arrow is implemented as the
 identity function that just returns back its argument. The

src/content/1.10/Natural Transformations.tex

@@ -86,15 +86,15 @@
 condition} that holds for any \code{f}:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-G f ◦ α\textsubscript{a} = α\textsubscript{b} ◦ F f
+G f  \ensuremath{\circ} α\textsubscript{a} = α\textsubscript{b}  \ensuremath{\circ} F f
 \end{Verbatim}
 The naturality condition is a pretty stringent requirement. For
 instance, if the morphism \code{F f} is invertible, naturality
 determines \code{α\textsubscript{b}} in terms of \code{α\textsubscript{a}}. It \newterm{transports}
 \code{α\textsubscript{a}} along \code{f}:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-α\textsubscript{b} = (G f) ◦ α\textsubscript{a} ◦ (F f)\textsuperscript{-1}
+α\textsubscript{b} = (G f)  \ensuremath{\circ} α\textsubscript{a}  \ensuremath{\circ} (F f)\textsuperscript{-1}
 \end{Verbatim}
 
 \begin{wrapfigure}[7]{R}{0pt}
@@ -201,7 +201,7 @@ \section{Polymorphic Functions}\label{polymorphic-functions}
 is a function \code{f::a->b}):
 
 \begin{Verbatim}[commandchars=\\\{\}]
-G f ◦ α\textsubscript{a} = α\textsubscript{b} ◦ F f
+G f  \ensuremath{\circ} α\textsubscript{a} = α\textsubscript{b}  \ensuremath{\circ} F f
 \end{Verbatim}
 In Haskell, the action of a functor \code{G} on a morphism \code{f}
 is implemented using \code{fmap}. I'll first write it in
@@ -476,13 +476,13 @@ \section{Functor Category}\label{functor-category}
 morphism:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-β\textsubscript{a} ◦ α\textsubscript{a} :: F a -> H a
+β\textsubscript{a}  \ensuremath{\circ} α\textsubscript{a} :: F a -> H a
 \end{Verbatim}
 We will use this morphism as the component of the natural transformation
 β ⋅ α --- the composition of two natural transformations β after α:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-(β \ensuremath{⋅} α)a = β\textsubscript{a} ◦ α\textsubscript{a}
+(β \ensuremath{⋅} α)a = β\textsubscript{a}  \ensuremath{\circ} α\textsubscript{a}
 \end{Verbatim}
 
 \begin{figure}[H]
@@ -495,7 +495,7 @@ \section{Functor Category}\label{functor-category}
 composition is indeed a natural transformation from F to H:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-H f ◦ (β \ensuremath{⋅} α)\textsubscript{a} = (β \ensuremath{⋅} α)\textsubscript{b} ◦ F f
+H f  \ensuremath{\circ} (β \ensuremath{⋅} α)\textsubscript{a} = (β \ensuremath{⋅} α)\textsubscript{b}  \ensuremath{\circ} F f
 \end{Verbatim}
 
 \begin{figure}[H]
@@ -611,7 +611,7 @@ \section{2-Categories}\label{categories}
 Instead of a Hom-set between two categories C and D, we have a
 Hom-category --- the functor category D\textsuperscript{C}. We have
 regular functor composition: a functor F from D\textsuperscript{C}
-composes with a functor G from E\textsuperscript{D} to give G ◦ F from
+composes with a functor G from E\textsuperscript{D} to give G  \ensuremath{\circ} F from
 E\textsuperscript{C}. But we also have composition inside each
 Hom-category --- vertical composition of natural transformations, or
 2-morphisms, between functors.
@@ -629,7 +629,7 @@ \section{2-Categories}\label{categories}
 and their composition:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-G ◦ F :: C -> E
+G  \ensuremath{\circ} F :: C -> E
 \end{Verbatim}
 Suppose we have two natural transformations, α and β, that act,
 respectively, on functors F and G:
@@ -650,10 +650,10 @@ \section{2-Categories}\label{categories}
 members of two different functor categories: D \textsuperscript{C} and E
 \textsuperscript{D}. We can, however, apply composition to the functors
 F' and G', because the target of F' is the source of G' --- it's the
-category D. What's the relation between the functors G'◦ F' and G ◦ F?
+category D. What's the relation between the functors G' \ensuremath{\circ} F' and G  \ensuremath{\circ} F?
 
 Having α and β at our disposal, can we define a natural transformation
-from G ◦ F to G'◦ F'? Let me sketch the construction.
+from G  \ensuremath{\circ} F to G' \ensuremath{\circ} F'? Let me sketch the construction.
 
 \begin{figure}[H]
 \centering
@@ -690,26 +690,26 @@ \section{2-Categories}\label{categories}
 \end{Verbatim}
 That's a lot of morphisms. Our goal is to find a morphism that goes from
 \code{G (F a)} to \code{G'(F'a)}, a candidate for the
-component of a natural transformation connecting the two functors G ◦ F
-and G'◦ F'. In fact there's not one but two paths we can take from
+component of a natural transformation connecting the two functors G  \ensuremath{\circ} F
+and G' \ensuremath{\circ} F'. In fact there's not one but two paths we can take from
 \code{G (F a)} to \code{G'(F'a)}:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-G'α\textsubscript{a} ◦ β\textsubscript{F a}
-β\textsubscript{F'a} ◦ G α\textsubscript{a}
+G'α\textsubscript{a}  \ensuremath{\circ} β\textsubscript{F a}
+β\textsubscript{F'a}  \ensuremath{\circ} G α\textsubscript{a}
 \end{Verbatim}
 Luckily for us, they are equal, because the square we have formed turns
 out to be the naturality square for β.
 
-We have just defined a component of a natural transformation from G ◦ F
-to G'◦ F'. The proof of naturality for this transformation is pretty
+We have just defined a component of a natural transformation from G  \ensuremath{\circ} F
+to G' \ensuremath{\circ} F'. The proof of naturality for this transformation is pretty
 straightforward, provided you have enough patience.
 
 We call this natural transformation the \newterm{horizontal composition} of
 α and β:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-β ◦ α :: G ◦ F -> G' ◦ F'
+β  \ensuremath{\circ} α :: G  \ensuremath{\circ} F -> G'  \ensuremath{\circ} F'
 \end{Verbatim}
 Again, following Mac Lane I use the small circle for horizontal
 composition, although you may also encounter star in its place.
@@ -746,7 +746,7 @@ \section{2-Categories}\label{categories}
 Finally, the two compositions satisfy the interchange law:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-(β' \ensuremath{\cdot} α') ◦ (β \ensuremath{\cdot} α) = (β' ◦ β) \ensuremath{\cdot} (α' ◦ α)
+(β' \ensuremath{\cdot} α')  \ensuremath{\circ} (β \ensuremath{\cdot} α) = (β'  \ensuremath{\circ} β) \ensuremath{\cdot} (α'  \ensuremath{\circ} α)
 \end{Verbatim}
 
 I will quote Saunders Mac Lane here: The reader may enjoy writing down
@@ -761,7 +761,7 @@ \section{2-Categories}\label{categories}
 notation:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-F ◦ α
+F  \ensuremath{\circ} α
 \end{Verbatim}
 where F is a functor from D to E, and α is a natural transformation
 between two functors going from C to D. Since you can't compose a
@@ -772,7 +772,7 @@ \section{2-Categories}\label{categories}
 Similarly:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-α ◦ F
+α  \ensuremath{\circ} F
 \end{Verbatim}
 is a horizontal composition of α after 1\textsubscript{F}.
 

src/content/1.3/Categories Great and Small.tex

@@ -297,7 +297,7 @@ \section{Monoid as Category}\label{monoid-as-category}
 element corresponding to the composition of the corresponding morphisms.
 If you give me two elements of \code{M(m, m)} corresponding to \code{f} and
 \code{g}, their product will correspond to the composition
-\code{g◦f}. The composition always exists, because the source and the
+\code{g \ensuremath{\circ} f}. The composition always exists, because the source and the
 target for these morphisms are the same object. And it's associative by
 the rules of category. The identity morphism is the neutral element of
 this product. So we can always recover a set monoid from a category

src/content/1.5/Products and Coproducts.tex

@@ -142,10 +142,10 @@ \section{Duality}\label{duality}
 all the requirements of a category, as long as we simultaneously
 redefine composition. If original morphisms
 \code{f::a->b} and \code{g::b->c} composed
-to \code{h::a->c} with \code{h=g◦f}, then the reversed
+to \code{h::a->c} with \code{h=g \ensuremath{\circ} f}, then the reversed
 morphisms \code{f\textsuperscript{op}::b->a} and
 \code{g\textsuperscript{op}::c->b} will compose to
-\code{h\textsuperscript{op}::c->a} with \code{h\textsuperscript{op}=f\textsuperscript{op}◦g\textsuperscript{op}}. And reversing
+\code{h\textsuperscript{op}::c->a} with \code{h\textsuperscript{op}=f\textsuperscript{op} \ensuremath{\circ} g\textsuperscript{op}}. And reversing
 the identity arrows is a (pun alert!) no-op.
 
 Duality is a very important property of categories because it doubles
@@ -211,13 +211,13 @@ \section{Isomorphisms}\label{isomorphisms}
 \end{figure}
 
 \noindent
-The composition \emph{g◦f} must be a morphism from i\textsubscript{1} to
+The composition \emph{g \ensuremath{\circ} f} must be a morphism from i\textsubscript{1} to
 i\textsubscript{1}. But i\textsubscript{1} is initial so there can only
 be one morphism going from i\textsubscript{1} to i\textsubscript{1}.
 Since we are in a category, we know that there is an identity morphism
 from i\textsubscript{1} to i\textsubscript{1}, and since there is room
-for only one, that must be it. Therefore \emph{g◦f} is equal to
-identity. Similarly, \emph{f◦g} must be equal to identity, because there
+for only one, that must be it. Therefore \emph{g \ensuremath{\circ} f} is equal to
+identity. Similarly, \emph{f \ensuremath{\circ} g} must be equal to identity, because there
 can be only one morphism from i\textsubscript{2} back to
 i\textsubscript{2}. This proves that \emph{f} and \emph{g} must be the
 inverse of each other. Therefore any two initial objects are isomorphic.

src/content/1.8/Functoriality.tex

@@ -32,7 +32,7 @@ \section{Bifunctors}\label{bifunctors}
 way:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-(f, g) ◦ (f', g') = (f ◦ f', g ◦ g')
+(f, g)  \ensuremath{\circ} (f', g') = (f  \ensuremath{\circ} f', g  \ensuremath{\circ} g')
 \end{Verbatim}
 The composition is associative and it has an identity --- a pair of
 identity morphisms \emph{(id, id)}. So a cartesian product of categories

src/content/1.9/Function Types.tex

@@ -144,7 +144,7 @@ \section{Universal Construction}\label{universal-construction}
 application \code{g'}:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-g' = g ◦ (h × id)
+g' = g  \ensuremath{\circ} (h × id)
 \end{Verbatim}
 The key here is the action of the product on morphisms.
 
@@ -192,7 +192,7 @@ \section{Universal Construction}\label{universal-construction}
 that factors \code{g} through \code{eval}:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-g = eval ◦ (h × id)
+g = eval  \ensuremath{\circ} (h × id)
 \end{Verbatim}
 \strut
 \end{minipage}\tabularnewline
@@ -244,7 +244,7 @@ \section{Currying}\label{currying}
 the two-argument function \code{g} using the formula:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-g = eval ◦ (h × id)
+g = eval  \ensuremath{\circ} (h × id)
 \end{Verbatim}
 The function \code{g} can be called the \newterm{uncurried} version of
 \code{h}.

src/content/2.4/Representable Functors.tex

@@ -87,7 +87,7 @@ \section{The Hom Functor}\label{the-hom-functor}
 \code{h} matches the source of \code{f}, so their composition:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-f ◦ h :: a -> y
+f  \ensuremath{\circ} h :: a -> y
 \end{Verbatim}
 is a morphism going from \code{a} to \code{y}. It is therefore a
 member of \code{C(a, y)}.
@@ -104,7 +104,7 @@ \section{The Hom Functor}\label{the-hom-functor}
 and its action on a morphism \code{h} as:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-C(a, f) h = f ◦ h
+C(a, f) h = f  \ensuremath{\circ} h
 \end{Verbatim}
 Since this construction works in any category, it must also work in the
 category of Haskell types. In Haskell, the hom-functor is better known
@@ -213,7 +213,7 @@ \section{Representable Functors}\label{representable-functors}
 commutes:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-F f ◦ α\textsubscript{x} = α\textsubscript{y} ◦ C(a, f)
+F f  \ensuremath{\circ} α\textsubscript{x} = α\textsubscript{y}  \ensuremath{\circ} C(a, f)
 \end{Verbatim}
 In Haskell, we would replace natural transformations with polymorphic
 functions:
@@ -245,7 +245,7 @@ \section{Representable Functors}\label{representable-functors}
 It must respect naturality conditions, and it must be the inverse of α:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-α ◦ β = id = β ◦ α
+α  \ensuremath{\circ} β = id = β  \ensuremath{\circ} α
 \end{Verbatim}
 We will see later that a natural transformation from \code{C(a, -)}
 to any \textbf{Set}-valued functor always exists (Yoneda's lemma) but it

src/content/2.5/The Yoneda Lemma.tex

@@ -68,7 +68,7 @@
 So the naturality square for \code{α}:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-α\textsubscript{y} ◦ C(a, f) = F f ◦ α\textsubscript{x}
+α\textsubscript{y}  \ensuremath{\circ} C(a, f) = F f  \ensuremath{\circ} α\textsubscript{x}
 \end{Verbatim}
 becomes, point-wise, when acting on \code{h}:
 
@@ -80,12 +80,12 @@
 precomposition:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-C(a, f) h = f ◦ h
+C(a, f) h = f  \ensuremath{\circ} h
 \end{Verbatim}
 which leads to:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-αy (f ◦ h) = (F f) (αx h)
+αy (f  \ensuremath{\circ} h) = (F f) (αx h)
 \end{Verbatim}
 Just how strong this condition is can be seen by specializing it to the
 case of \code{x} equal to \code{a}.
@@ -178,7 +178,7 @@
 \noindent
 Now let's consider the action of \code{C(a, g)} on our original
 \code{p} which, as you remember, corresponds to \code{id\textsubscript{a}}. It is
-defined as precomposition, \code{g◦id\textsubscript{a}}, which is equal to \code{g},
+defined as precomposition, \code{g \ensuremath{\circ} id\textsubscript{a}}, which is equal to \code{g},
 which corresponds to our point \code{p'}. So the morphism
 \code{g} is mapped to a function that, when acting on \code{p}
 produces \code{p'}, which is \code{g}. We have come full
@@ -206,7 +206,7 @@
 F g :: F a -> F x
 \end{Verbatim}
 Again, \code{C(a, g)} acting on our \code{p} is given by the
-precomposition: \code{g ◦ id\textsubscript{a}}, which corresponds to a point
+precomposition: \code{g  \ensuremath{\circ} id\textsubscript{a}}, which corresponds to a point
 \code{p'} in \code{C(a, x)}. Naturality determines the value
 of \code{αx} acting on \code{p'} to be:
 

src/content/2.6/Yoneda Embedding.tex

@@ -298,12 +298,12 @@ \section{Naturality}\label{naturality}
 \code{f} lifted by \code{G}:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-(G f) ◦ Φ\textsubscript{a}
+(G f)  \ensuremath{\circ} Φ\textsubscript{a}
 \end{Verbatim}
 Notice that, because of naturality of \code{Φ}, this is the same as:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-Φ\textsubscript{b} ◦ (F f)
+Φ\textsubscript{b}  \ensuremath{\circ} (F f)
 \end{Verbatim}
 I'm not going to prove the naturality of the whole isomorphism --- after
 you've established what the functors are, the proof is pretty

src/content/3.1/It's All About Morphisms.tex

@@ -134,7 +134,7 @@ \section{Hom-Sets}\label{hom-sets}
 with \code{f} is different than that with \code{g}, for instance:
 
 \begin{Verbatim}[commandchars=\\\{\}]
-h ◦ f ≠ h ◦ g
+h  \ensuremath{\circ} f ≠ h  \ensuremath{\circ} g
 \end{Verbatim}
 then we can directly ``observe'' the difference between \code{f} and
 \code{g}. But even if the difference is not directly observable, we
@@ -148,7 +148,7 @@ \section{Hom-Sets}\label{hom-sets}
 resolution, e.g.,
 
 \begin{Verbatim}[commandchars=\\\{\}]
-h' ◦ F f ≠ h' ◦ F g
+h'  \ensuremath{\circ} F f ≠ h'  \ensuremath{\circ} F g
 \end{Verbatim}
 where \code{h'} is not in the image of \code{F}.