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smash.tex: small changes, add some preliminary references

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Notes/smash.tex
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@ -81,7 +81,7 @@ We define the pointed equivalences:
with underlying map defined with $\twist(f) \defeq \lam{b}\lam{a}f(a)(b)$. with underlying map defined with $\twist(f) \defeq \lam{b}\lam{a}f(a)(b)$.
\end{defn} \end{defn}
\begin{lem} \begin{lem}\label{lem:composition-pointed}
Given maps $f:A'\pmap A$ and $g:B\pmap B'$. Then there are maps Given maps $f:A'\pmap A$ and $g:B\pmap B'$. Then there are maps
$(f\pmap C):(A\pmap C)\pmap(A'\pmap C)$ and $(C\pmap g):(C\pmap B)\pmap(C\pmap B')$ given by $(f\pmap C):(A\pmap C)\pmap(A'\pmap C)$ and $(C\pmap g):(C\pmap B)\pmap(C\pmap B')$ given by
precomposition with $f$, resp. postcomposition with $g$. The map $\lam{g}C\pmap g$ preserves the basepoint, giving rise to a map $$(C\pmap ({-})):(B\pmap B')\pmap(C\pmap B)\pmap(C\pmap B').$$ precomposition with $f$, resp. postcomposition with $g$. The map $\lam{g}C\pmap g$ preserves the basepoint, giving rise to a map $$(C\pmap ({-})):(B\pmap B')\pmap(C\pmap B)\pmap(C\pmap B').$$
@ -98,34 +98,34 @@ We define the pointed equivalences:
\section{Naturality and a version of the Yoneda lemma} \section{Naturality and a version of the Yoneda lemma}
\begin{defn}\label{def:naturality} \begin{defn}\label{def:naturality}
Let $F$, $G$ be functors of pointed types, i.e. pointed maps with a functorial action (e.g. if $f : A \to B$, then we can define $F(f) : F(A) \to F(B)$, respecting identity and composition). A (1-coherent) \emph{functor} $F$ between pointed types is a function $F_0:\type^*\to\type^*$ with an action of morphisms $F_1 : (A \to^* B) \to (FA \to^* FB)$ such that $F_1(g \o f)\sim F_1g \o F_1 f$ and $F_1\idfunc[A]\sim\idfunc[F_0A]$. We will write both $F_0$ and $F_1$ as $F$. A functor $F$ is \emph{pointed} if $F\unit$ is contractible. In this case $F0_{A,B}\sim 0_{FA,FB}$.
Let $\theta : F \Rightarrow G$ be a natural transformation from $F$ to $G$, i.e. a pointed map $F(X) \to G(X)$ for all pointed types $X$. For every $f : A \to B$, there is a diagram: Let $F$, $G$ be functors of pointed types. Suppose that we have a transformation $\theta_X : F(X) \to G(X)$ for all pointed types $X$.
\begin{center} % \begin{center}
\begin{tikzcd} % \begin{tikzcd}
F(A) % F(A)
\arrow[r, "F(f)"] % \arrow[r, "F(f)"]
\arrow[d, swap, "\theta_A"] % \arrow[d, swap, "\theta_A"]
& F(B) % & F(B)
\arrow[d, "\theta_B"] % \arrow[d, "\theta_B"]
\\ % \\
G(A) % G(A)
\arrow[r, swap, "G(f)"] % \arrow[r, swap, "G(f)"]
& G(B) % & G(B)
\end{tikzcd} % \end{tikzcd}
\end{center} % \end{center}
We define the following notions of naturality for $\theta$: We define the following notions of naturality for $\theta$:
\begin{itemize} \begin{itemize}
\item \textbf{(strong) naturality} will refer to a pointed homotopy \item \textbf{naturality} will refer to a pointed homotopy
\[p_\theta(f) : G(f) \o \theta_A \sim \theta_B \o F(f)\] \[p_\theta(f) : G(f) \o \theta_A \sim \theta_B \o F(f)\]
for every $f : A \to B$ and \textbf{weak naturality} to the underlying (non-pointed) homotopy; for every $f : A \to B$ and \textbf{weak naturality} to the underlying (non-pointed) homotopy;
\item \textbf{pointed (strong) naturality} will refer to the same pointed homotopy, with the additional condition that $p_\theta(0) = (p_\theta)_0$, where \item \textbf{pointed naturality} will refer to the same pointed homotopy, with the additional condition that $p_\theta(0) = (p_\theta)_0$, where
\[(p_\theta)_0 : G(0) \o \theta_A \sim 0 \o \theta_A \sim 0 \sim \theta_B \o 0 \sim \theta_B \o F(0)\] \[(p_\theta)_0 : G(0) \o \theta_A \sim 0 \o \theta_A \sim 0 \sim \theta_B \o 0 \sim \theta_B \o F(0)\]
is the canonical proof of the pointed homotopy $G(0) \o \theta_A \sim \theta_B \o F(0)$, whereas \textbf{pointed weak naturality} will refer to the corresponding non-pointed condition. is the canonical proof of the pointed homotopy $G(0) \o \theta_A \sim \theta_B \o F(0)$, whereas \textbf{pointed weak naturality} will refer to the corresponding non-pointed condition.
\end{itemize} \end{itemize}
\end{defn} \end{defn}
\begin{rmk} \begin{rmk}
The relation between the four notions of naturality is as expected: strong implies weak, and pointed implies simple. Weak naturality is generally ill-behaved: for example, weak naturality of $\theta$ does not imply weak naturality of $\theta \to X$ or $X \to \theta$, whereas the implication holds for strong naturality. The relation between the four notions of naturality is as expected: naturality implies weak naturality, and pointed implies simple. Weak naturality is generally ill-behaved: for example, weak naturality of $\theta$ does not imply weak naturality of $\theta \to X$ or $X \to \theta$, whereas the implication holds for naturality.
\end{rmk} \end{rmk}
\begin{rmk} \begin{rmk}
@ -133,7 +133,7 @@ We define the pointed equivalences:
\end{rmk} \end{rmk}
\begin{lem}[Yoneda]\label{lem:yoneda} \begin{lem}[Yoneda]\label{lem:yoneda}
Let $A$, $B$ be pointed types, and assume, for all pointed types $X$, a pointed equivalence $\phi_X : (B \to X) \simeq (A \to X)$, natural in $X$, i.e. for all $f : X \to X'$ there is a homotopy \[ p_\phi(f) : (A \to f) \o \phi_X \sim \phi_X' \o (B \to f) \] Let $A$, $B$ be pointed types, and assume, for all pointed types $X$, a pointed equivalence $\phi_X : (B \to X) \simeq (A \to X)$, natural in $X$, i.e. for all $f : X \to X'$ there is a pointed homotopy \[ p_\phi(f) : (A \to f) \o \phi_X \sim \phi_{X'} \o (B \to f) \]
% making the following diagram commute for all $f : X \to X'$: % making the following diagram commute for all $f : X \to X'$:
% \begin{center} % \begin{center}
% \begin{tikzcd} % \begin{tikzcd}
@ -151,11 +151,11 @@ We define the pointed equivalences:
Then there exists a pointed equivalence $\psi_\phi : A \simeq B$. Then there exists a pointed equivalence $\psi_\phi : A \simeq B$.
\end{lem} \end{lem}
\begin{proof} \begin{proof}
We define $\psi_\phi \defeq \phi_B(\idfunc[B]) : A \to B$ and $\psi_\phi\sy \defeq \phi_A\sy(\idfunc[A])$. The given naturality square for $X \defeq B$ and $g \defeq \psi_\phi\sy$ yields $\psi_\phi\sy \o \phi_B (\idfunc[B]) \judgeq \psi_\phi\sy \o \psi_\phi \sim \phi_A (\psi_\phi\sy \o \idfunc[B]) \judgeq \phi_A (\phi_A\sy (\idfunc[A])) \sim \idfunc[A]$, and similarly for the inverse composition. We define $\psi_\phi \defeq \phi_B(\idfunc[B]) : A \to B$ and $\psi_\phi\sy \defeq \phi_A\sy(\idfunc[A]):B\to A$. The given naturality square for $f \defeq \psi_\phi\sy$ yields $\psi_\phi\sy \o \phi_B (\idfunc[B]) \judgeq \psi_\phi\sy \o \psi_\phi \sim \phi_A (\psi_\phi\sy \o \idfunc[B]) \judgeq \phi_A (\phi_A\sy (\idfunc[A])) \sim \idfunc[A]$, and similarly for the inverse composition.
\end{proof} \end{proof}
\begin{lem}\label{lem:yoneda-pointed} \begin{lem}\label{lem:yoneda-pointed}
Assume $A$, $B$, $\phi_X$ and $p$ as in \autoref{lem:yoneda}, and assume moreover that $\phi_X$ is pointed natural. Then there is a pointed homotopy $(\psi_\phi \to X) \sim \phi_X$. Assume $A$, $B$, $\phi_X$ and $p$ as in \autoref{lem:yoneda}, and assume moreover that $\phi$ is pointed natural. Then there is a pointed homotopy $(\psi_\phi \to X) \sim \phi_X$.
\end{lem} \end{lem}
\begin{proof} \begin{proof}
@ -178,7 +178,7 @@ We define the pointed equivalences:
\end{tikzcd} \end{tikzcd}
\end{center} \end{center}
where the top-left expression is definitionally equal to $0 \o \phi_X(\idfunc)$, the horizontal path comes from the underlying homotopy and $(\phi_X)_0$ is the canonical path from $\phi_X(0)$ to $0$. Since $\phi_X$ is pointed natural, we have that where the top-left expression is definitionally equal to $0 \o \phi_X(\idfunc)$, the horizontal path comes from the underlying homotopy and $(\phi_X)_0$ is the canonical path from $\phi_X(0)$ to $0$. Since $\phi_X$ is pointed natural, we have that
$p_{\phi_X}(0)(\idfunc) = (p_{\phi_X})_0(\idfunc)$, which, in this case, is: $p_{\phi_X}(0)(\idfunc) = (p_{\phi_X})_0(\idfunc)$, which, in this case, is the concatenation:
\begin{align*} \begin{align*}
0\o \phi_X(\idfunc) 0\o \phi_X(\idfunc)
&= 0 &&\text{(by $\zeroh_{q_X(\idfunc)}$)}\\ &= 0 &&\text{(by $\zeroh_{q_X(\idfunc)}$)}\\
@ -625,11 +625,11 @@ square, which follows from the left pentagon in \autoref{lem:smash-coh}.
0 0
\end{tikzcd} \end{tikzcd}
\end{center} \end{center}
To show that this naturality is pointed, we need to show that if $g=0$ then this homotopy is the same as the concatenation of the following pointed homotopies: To show that this naturality is pointed, we need to show that if $g=0$ then this homotopy is the same as the concatenation $q_0$ of the following pointed homotopies:
$$q:({-})\smsh C \circ (A \to 0)\sim ({-})\smsh C \circ 0 \sim 0 \sim 0 \circ ({-})\smsh C\sim 0\smsh B \circ ({-})\smsh C.$$ $$({-})\smsh C \circ (A \to 0)\sim ({-})\smsh C \circ 0 \sim 0 \sim 0 \circ ({-})\smsh C\sim 0\smsh B \circ ({-})\smsh C.$$
To show that the underlying homotopies are the same, we need to show that $i(0,f,\idfunc[C],\idfunc[C])$ is equal to the following concatenation of pointed homotopies To show that the underlying homotopies are the same, we need to show that $i(0,f,\idfunc[C],\idfunc[C])$ is equal to the following concatenation of pointed homotopies
$$q(f):(0\circ f)\smsh C\sim 0\smsh C \sim 0 \sim 0 \circ f\smsh C\sim 0\smsh B \circ f\smsh C,$$ $$q_0(f):(0\circ f)\smsh C\sim 0\smsh C \sim 0 \sim 0 \circ f\smsh C\sim 0\smsh B \circ f\smsh C,$$
which is the right pentagons in \autoref{lem:smash-coh}. which is the right pentagon in \autoref{lem:smash-coh}.
To show that these pointed homotopies respect the basepoint in the same way, we need to show that (TODO) To show that these pointed homotopies respect the basepoint in the same way, we need to show that (TODO)
``$R\mathrel\square(0\smsh C \circ t)\cdot q_0=L$ where $L$ and $R$ are the left and right pentagons applied to $0$ and $\square$ is whiskering.'' ``$R\mathrel\square(0\smsh C \circ t)\cdot q_0=L$ where $L$ and $R$ are the left and right pentagons applied to $0$ and $\square$ is whiskering.''
@ -664,7 +664,7 @@ are filled by (corollaries of) \autoref{lem:smash-general}.
$\epsilon_{B,C}\equiv\epsilon : (B\pmap C)\smsh B \pmap C$ dinatural in $B$ and pointed natural in $C$. $\epsilon_{B,C}\equiv\epsilon : (B\pmap C)\smsh B \pmap C$ dinatural in $B$ and pointed natural in $C$.
These maps satisfy the unit-counit laws: These maps satisfy the unit-counit laws:
$$(A\to\epsilon_{A,B})\o \eta_{A\to B,A}\sim \idfunc[A\to B]\qquad $$(A\to\epsilon_{A,B})\o \eta_{A\to B,A}\sim \idfunc[A\to B]\qquad
\epsilon_{B,B\smsh C}\o \eta_{A,B}\smsh B\sim\idfunc[A\smsh B].$$ \epsilon_{B,A\smsh B}\o \eta_{A,B}\smsh B\sim\idfunc[A\smsh B].$$
\end{lem} \end{lem}
Note: $\eta$ is also dinatural in $B$, but we don't need this. Note: $\eta$ is also dinatural in $B$, but we don't need this.
\begin{proof} \begin{proof}
@ -702,7 +702,7 @@ Note: $\eta$ is also dinatural in $B$, but we don't need this.
then we need to show that $f(b_0)=c_0$, which is true by $f_0$. If $x$ varies over $\gluer_b$ we then we need to show that $f(b_0)=c_0$, which is true by $f_0$. If $x$ varies over $\gluer_b$ we
need to show that $0(b)=c_0$ which is true by reflexivity. Now $\epsilon_0\defeq 1:\epsilon(0_{B,C},b_0)=c_0$ shows that $\epsilon$ is pointed. need to show that $0(b)=c_0$ which is true by reflexivity. Now $\epsilon_0\defeq 1:\epsilon(0_{B,C},b_0)=c_0$ shows that $\epsilon$ is pointed.
Now we need to show that the counit is natural in $B$ and pointed natural in $C$. Now we need to show that the counit is dinatural in $B$ and pointed natural in $C$. (TODO)
Finally, we need to show the unit-counit laws. For the underlying homotopy of the first one, let Finally, we need to show the unit-counit laws. For the underlying homotopy of the first one, let
$f:A\to B$. We need to show that $p_f:\epsilon\o\eta f\sim f$. We define $p_f(a)=1:\epsilon(f,a)=f(a)$. To show that $p_f$ is a pointed homotopy, we need to show that $f:A\to B$. We need to show that $p_f:\epsilon\o\eta f\sim f$. We define $p_f(a)=1:\epsilon(f,a)=f(a)$. To show that $p_f$ is a pointed homotopy, we need to show that
@ -874,7 +874,7 @@ Using \autoref{lem:yoneda} (Yoneda) we can prove associativity, left- and right
\end{proof} \end{proof}
\begin{thm}[Associativity pentagon]\label{thm:smash-associativity-pentagon} \begin{thm}[Associativity pentagon]\label{thm:smash-associativity-pentagon}
For $A$, $B$, $C$ and $D$ pointed types, there is a homotopy For $A$, $B$, $C$ and $D$ pointed types, there is a pointed homotopy
\[\alpha \o \alpha \sim (A \smsh \alpha) \o \alpha \o (\alpha \smsh D)\] \[\alpha \o \alpha \sim (A \smsh \alpha) \o \alpha \o (\alpha \smsh D)\]
corresponding to the commutativity of the following diagram: corresponding to the commutativity of the following diagram:
\begin{center} \begin{center}
@ -1013,7 +1013,7 @@ Using \autoref{lem:yoneda} (Yoneda) we can prove associativity, left- and right
\end{proof} \end{proof}
\begin{thm}[Unitors triangle]\label{thm:smash-unitors-triangle} \begin{thm}[Unitors triangle]\label{thm:smash-unitors-triangle}
For $A$ and $B$ pointed types, there is a homotopy For $A$ and $B$ pointed types, there is a pointed homotopy
\[(A \smsh \lambda) \o \alpha \sim (\rho \smsh B)\] \[(A \smsh \lambda) \o \alpha \sim (\rho \smsh B)\]
corresponding to the commutativity of the following diagram: corresponding to the commutativity of the following diagram:
\begin{center} \begin{center}
@ -1045,7 +1045,7 @@ Using \autoref{lem:yoneda} (Yoneda) we can prove associativity, left- and right
\end{proof} \end{proof}
\begin{thm}[Braiding-unitors triangle]\label{thm:smash-braiding-unitors} \begin{thm}[Braiding-unitors triangle]\label{thm:smash-braiding-unitors}
For a pointed type $A$, there is a homotopy For a pointed type $A$, there is a pointed homotopy
\[\lambda \o \gamma \sim \rho\] \[\lambda \o \gamma \sim \rho\]
corresponding to the commutativity of the following diagram: corresponding to the commutativity of the following diagram:
\begin{center} \begin{center}
@ -1121,12 +1121,12 @@ Using \autoref{lem:yoneda} (Yoneda) we can prove associativity, left- and right
& (A \to B \to C \to X) & (A \to B \to C \to X)
\end{tikzcd} \end{tikzcd}
\end{center} \end{center}
where the squares on the right are instances of naturality of $\twist$, while the commutativity of the pentagon on the left follows easily from the definition of $\twist$. where the square on the top right is by \autoref{lem:composition-pointed}, the square on the bottom right is naturality of $\twist$. The commutativity of the pentagon on the left is TODO (it is a diagram of a 1-coherent symmetric closed category).
\end{proof} \end{proof}
\begin{thm}[Associativity-braiding hexagon]\label{thm:smash-associativity-braiding} \begin{thm}[Associativity-braiding hexagon]\label{thm:smash-associativity-braiding}
For pointed types $A$, $B$ and $C$, there is a homotopy For pointed types $A$, $B$ and $C$, there is a pointed homotopy
\[\alpha \o \gamma \o \alpha \sim (B \smsh \gamma) \o \alpha \o (\gamma \smsh C)\] \[\alpha \o \gamma \o \alpha \sim (B \smsh \gamma) \o \alpha \o (\gamma \smsh C)\]
corresponding to the commutativity of the following diagram: corresponding to the commutativity of the following diagram:
\begin{center} \begin{center}
@ -1178,7 +1178,7 @@ Using \autoref{lem:yoneda} (Yoneda) we can prove associativity, left- and right
\end{proof} \end{proof}
\begin{thm}[Double braiding]\label{thm:smash-double-braiding} \begin{thm}[Double braiding]\label{thm:smash-double-braiding}
For $A$ and $B$ pointed types, there is a homotopy For $A$ and $B$ pointed types, there is a pointed homotopy
\[\gamma \o \gamma \sim \idfunc\] \[\gamma \o \gamma \sim \idfunc\]
corresponding to the commutativity of the following diagram: corresponding to the commutativity of the following diagram:
\begin{center} \begin{center}
@ -1232,6 +1232,15 @@ If $F$ is a pointed 2-coherent functor (or more precisely a 1-coherent functor w
Now we can also show \autoref{thm:smash-functor-right} more generally, but we will only formulate that for functors between pointed types. If $F:\type^*\to\type^*$ is a 2-coherent pointed functor, then it induces a map Now we can also show \autoref{thm:smash-functor-right} more generally, but we will only formulate that for functors between pointed types. If $F:\type^*\to\type^*$ is a 2-coherent pointed functor, then it induces a map
$(A\to B)\to(FA\to FB)$ which is natural in $A$ and $B$. Moreover, if $F$ is 3-coherent then this is a pointed natural transformation in $B$. $(A\to B)\to(FA\to FB)$ which is natural in $A$ and $B$. Moreover, if $F$ is 3-coherent then this is a pointed natural transformation in $B$.
\section{References}
\begin{itemize}
\item (TODO)
\item An algebraic theory of tricategories, Nick Gurski
\item Closed Categories, Samuel Eilenberg and G. Max Kelly (Chapter 2, Theorem 5.3, 1-categorical account of getting a monoidal category from a enriched adjunction in a closed category)
\item Permutative categories, multicategories, and algebraic K-theory, A D Elmendorf and M A Mandell (Lemma 4.20 gives a 1-categorical account that if C is a symmetric monoidal closed bicomplete 1-category, then the category of pointed objects is a symmetric monoidal closed bicomplete 1-category).
\item On embedding closed categories, B.J. Day and M.L. Laplaza (definition of symmetric closed category).
\item Maybe: Handbook of Categorical Algebra 2 Categories and Structures, F Borceux. (Bjorn used it to look things up about symmetric monoidal closed categories)
\end{itemize}
\end{document} \end{document}