Small edits

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

iii/cat/03_adjunctions.tex

@@ -15,7 +15,7 @@ \subsection{Definition and examples}
         \[ \mathbf{Top}(DX, Y) \leftrightarrow \mathbf{Set}(X, UY) \]
         It also has a right adjoint \( I : \mathbf{Set} \to \mathbf{Top} \) which equips each set with its indiscrete topology.
         \[ \mathbf{Set}(UX, Y) \leftrightarrow \mathbf{Top}(X, IY) \]
-        \item Consider the functor \( \ob : \mathbf{Cat} \to \mathbf{Set} \) which maps each category to each set of objects.
+        \item Consider the functor \( \ob : \mathbf{Cat} \to \mathbf{Set} \) which maps each category to its set of objects.
         It has a left adjoint \( D \) which turns each set \( X \) into a discrete category in which the objects are elements of \( X \), and the only morphisms are identities.
         It also has a right adjoint \( I \) which turns each set \( X \) into an indiscrete category in which the objects are elements of \( X \), and there is exactly one morphism between any two elements of \( X \).
         In addition, \( D : \mathbf{Set} \to \mathbf{Cat} \) has a left adjoint \( \pi_0 : \mathbf{Cat} \to \mathbf{Set} \), where \( \pi_0 \mathcal C \) is the set of connected components of \( \ob \mathcal C \) under the graph induced by its morphisms.

iii/cat/06_monoidal_and_enriched_categories.tex

@@ -59,7 +59,7 @@ \subsection{Monoidal categories}
 	\arrow["{1_A \otimes \rho_B}", from=1-2, to=2-2]
 	\arrow["{\rho_{A \otimes B}}"', from=1-1, to=2-2]
 \end{tikzcd}\]
-Note that in the category of abelian groups with the usual tensor product, the obvious choice for \( \alpha \) is the map sending \( (a \otimes b) \otimes c \) to \( a \otimes (b \otimes c) \).
+Note that in the category of abelian groups with the usual tensor product, the obvious choice for \( \alpha_{A,B,C} \) is the map sending \( (a \otimes b) \otimes c \) to \( a \otimes (b \otimes c) \).
 However, there is also a natural isomorphism sending \( (a \otimes b) \otimes c \) to \( -a \otimes (b \otimes c) \).
 But this choice does not satisfy the pentagon equation, as a pentagon has an odd number of sides.