From dd14277e0bfc6270a488eb3b9ec231484065b9d8 Mon Sep 17 00:00:00 2001 From: spiceghello Date: Mon, 26 Mar 2018 09:11:55 +0200 Subject: [PATCH] additions on pentagons with interchange --- Notes/smash.tex | 167 +++++++++++++++++++++++++++++++++--------------- 1 file changed, 116 insertions(+), 51 deletions(-) diff --git a/Notes/smash.tex b/Notes/smash.tex index d0dc55a..2855ede 100644 --- a/Notes/smash.tex +++ b/Notes/smash.tex @@ -357,49 +357,59 @@ We define the pointed equivalences: Suppose that we have maps $A_1\lpmap{f_1}A_2\lpmap{f_2}A_3$ and $B_1\lpmap{g_1}B_2\lpmap{g_2}B_3$ and suppose that either $f_1$ or $f_2$ is constant. Then there are two homotopies $(f_2 \o f_1)\smsh (g_2 \o g_1)\sim 0$, one which uses the interchange law and one which does not. These two homotopies are equal. Specifically, the following two diagrams commute: - \begin{center} - \begin{tikzcd} + \begin{equation}\label{eq:pent-statement} + \begin{tikzcd}[column sep=2em] (f_2 \o 0)\smsh (g_2 \o g_1) \arrow[r, equals, "i"] \arrow[dd, swap, equals, "\zeroh' \smsh (g_2 \o g_1)"] &(f_2 \smsh g_2)\o (0 \smsh g_1) \arrow[d, equals, "(f_2 \smsh g_2) \o t_{g_1}"] - \\ - & (f_2 \smsh g_2)\o 0 - \arrow[d,equals, "\zeroh'"] - \\ - 0\smsh (g_2 \o g_1) - \arrow[r,equals, swap, "t_{g_2 \o g_1}"] - & 0 - \end{tikzcd} - \qquad - \begin{tikzcd} - (0 \o f_1)\smsh (g_2 \o g_1) + & (0 \o f_1)\smsh (g_2 \o g_1) \arrow[r, equals, "i"] \arrow[dd, swap, equals, "\zeroh \smsh (g_2 \o g_1)"] & (0 \smsh g_2)\o (f_1 \smsh g_1) \arrow[d,equals, "t_{g_2} \o (f_1 \smsh g_1)"] \\ - & 0\o (f_1 \smsh g_1) + & (f_2 \smsh g_2)\o 0 + \arrow[d,equals, "\zeroh'"] + && 0\o (f_1 \smsh g_1) \arrow[d,equals, "\zeroh"] \\ 0\smsh (g_2 \o g_1) + \arrow[r,equals, swap, "t_{g_2 \o g_1}"] + & 0 + & 0\smsh (g_2 \o g_1) \arrow[r,swap, equals, "t_{g_2 \o g_1}"] & 0 \end{tikzcd} - \end{center} +% \qquad +% \begin{tikzcd} +% (0 \o f_1)\smsh (g_2 \o g_1) +% \arrow[r, equals, "i"] +% \arrow[dd, swap, equals, "\zeroh \smsh (g_2 \o g_1)"] +% & (0 \smsh g_2)\o (f_1 \smsh g_1) +% \arrow[d,equals, "t_{g_2} \o (f_1 \smsh g_1)"] +% \\ +% & 0\o (f_1 \smsh g_1) +% \arrow[d,equals, "\zeroh"] +% \\ +% 0\smsh (g_2 \o g_1) +% \arrow[r,swap, equals, "t_{g_2 \o g_1}"] +% & 0 +% \end{tikzcd} + \end{equation} \end{lem} \begin{proof} % We will only do the case where $f_1\jdeq 0$, i.e. fill the diagram on the left. The other case is similar (and slightly easier). - \textbf{Case $f_1\judgeq 0$ (diagram on the left)}. First apply induction on the paths that $f_2$, $g_1$ and $g_2$ + \textbf{Case $f_1\judgeq 0$}. In order to fill the diagram on the left in (\ref{eq:pent-statement}), we first apply induction on the paths that $f_2$, $g_1$ and $g_2$ respect the basepoint. In this case $f_2\o0$ is definitionally equal to $0$, and the canonical proof that $f_2\o 0\sim0$ is (definitionally) equal to reflexivity. This means that the homotopy $(f_2 \o 0)\smsh (g_2 \o g_1)\sim0\smsh (g_2 \o g_1)$ is also equal to reflexivity, and also the path that $f_2 \smsh g_2$ respects the basepoint is reflexivity, hence the homotopy $(f_2 \smsh g_2)\o 0\sim0$ is also reflexivity. This means we need to fill the following square: - \begin{center} + \begin{equation}\label{eq:pent-left-all} \begin{tikzcd} (f_2 \o 0)\smsh (g_2 \o g_1) \arrow[r, equals,"i"] @@ -411,7 +421,7 @@ We define the pointed equivalences: \arrow[r, swap, equals,"t_{g_1 \o g_2}"] & 0 \end{tikzcd} - \end{center} + \end{equation} For the underlying homotopy, take $x : A_1\smsh B_1$ and apply induction on $x$. Suppose $x\equiv(a,b)$ for $a:A_1$ and $b:B_1$. With the notational convention for basepoints as in \autoref{lem:interchange}, we have to fill the square (we use that the paths that the maps respect the basepoints are reflexivity): \begin{equation}\label{eq:pent-left-ab} @@ -477,7 +487,35 @@ We define the pointed equivalences: \arrow[from=uu, equals, crossing over, very near start, "\mapfunc{f_2 \smsh g_2}(\gluer\tr\gluer\sy)"] \end{tikzcd} \end{equation} - Similarly, if $x$ varies over $\gluer_b$, we need to fill the cube below: the front and the back are the squares in (\ref{eq:pent-left-ab}) for $(a_1,b)$ and (\ref{eq:pent-left-auxr}) respectively; the left square is again degenerate; the other three sides come from the fact that $i$ and $t$ respect $\gluer_b$ (given in (\ref{eq:i-gluer}) and (\ref{eq:t-gluer}) respectively). Again, we omit the arguments of $\gluer$ in $\gluer\tr\gluer\sy$ (in this case, not a priori judgementally equal). + Such a cube can be generalized to the cube: + \begin{center} + \begin{tikzcd}[column sep=3em] + & h(y) + \arrow[rr, equals,"1"] + \arrow[dd, swap, equals, near end,"1"] + && h(y) + \arrow[dd,equals,"\mapfunc{h}(p_l\sy)"] + \\ + h(x) + \arrow[rr, equals, near end, crossing over, "1"] + \arrow[dd, swap, equals, "1"] + \arrow[ur, equals, "q_l"] + && h(x) + \arrow[ur, equals, near start, "\mapfunc{h}(p_l)"] + \\ + & h(y) + \arrow[rr, swap, equals, near start, "q_l\sy"] + && h(x) + \\ + h(x) + \arrow[rr, swap, equals,"q_r\tr q_r\sy"] + \arrow[ur, equals, "q_l"] + && h(x) + \arrow[from=uu, equals, crossing over, near start, "\mapfunc{h}(p_r\tr p_r\sy)"] + \arrow[ur, swap, equals, "1"] + \end{tikzcd} + \end{center} + for $X$ and $X'$ pointed types; a map $h : X \to X'$; terms $x$, $y$ $z : X$; paths $p_l : x = y$, $p_r : x = z$, $q_l : h(x) = h(y)$, $q_r : h(x) = h(z)$; and 2-paths $s_l : \mapfunc{h}(p_l) = q_l$ (for the back and the top) and $s_r : \mapfunc{h}(p_r) = q_r$ (for the right side). This cube is filled by path induction on $s_l$, $s_r$, $p_l$ and $p_r$. Finally, if $x$ varies over $\gluer_b$, we need to fill the cube below: the front and the back are the squares in (\ref{eq:pent-left-ab}) for $(a_1,b)$ and (\ref{eq:pent-left-auxr}) respectively; the left square is again degenerate; the other three sides come from the fact that $i$ and $t$ respect $\gluer_b$ (given in (\ref{eq:i-gluer}) and (\ref{eq:t-gluer}) respectively). Again, we omit the arguments of $\gluer$ in $\gluer\tr\gluer\sy$ (in this case, not a priori judgementally equal). \begin{equation}\label{eq:pent-left-gluer} \begin{tikzcd}[column sep=4em] & \auxr @@ -506,35 +544,7 @@ We define the pointed equivalences: \end{tikzcd} \end{equation} %After canceling applications of $\mapfunc{h\smsh k}(\gluer_z)=\gluer_{k(z)}$ on various sides of the squares (TODO). - In order to fill the cubes in (\ref{eq:pent-left-gluel}) and (\ref{eq:pent-left-gluer}), we generalize the paths and fill the cubes by path induction. The cube in (\ref{eq:pent-left-gluel}) can be generalized to a cube: - \begin{center} - \begin{tikzcd}[column sep=3em] - & h(y) - \arrow[rr, equals,"1"] - \arrow[dd, swap, equals, near end,"1"] - && h(y) - \arrow[dd,equals,"\mapfunc{h}(p_l\sy)"] - \\ - h(x) - \arrow[rr, equals, near end, crossing over, "1"] - \arrow[dd, swap, equals, "1"] - \arrow[ur, equals, "q_l"] - && h(x) - \arrow[ur, equals, near start, "\mapfunc{h}(p_l)"] - \\ - & h(y) - \arrow[rr, swap, equals, near start, "q_l\sy"] - && h(x) - \\ - h(x) - \arrow[rr, swap, equals,"q_r\tr q_r\sy"] - \arrow[ur, equals, "q_l"] - && h(x) - \arrow[from=uu, equals, crossing over, near start, "\mapfunc{h}(p_r\tr p_r\sy)"] - \arrow[ur, swap, equals, "1"] - \end{tikzcd} - \end{center} - for $X$ and $X'$ pointed types; a map $h : X \to X'$; terms $x$, $y$ $z : X$; paths $p_l : x = y$, $p_r : x = z$, $q_l : h(x) = h(y)$, $q_r : h(x) = h(z)$; and 2-paths $s_l : \mapfunc{h}(p_l) = q_l$ (for the back and the top) and $s_r : \mapfunc{h}(p_r) = q_r$ (for the right side). This cube is filled by path induction on $s_l$, $s_r$, $p_l$ and $p_r$. The cube in (\ref{eq:pent-left-gluer}) can be generalized to a similar cube: + This cube can be generalized to the cube: \begin{center} \begin{tikzcd}[column sep=3em] & h(y) @@ -562,14 +572,69 @@ We define the pointed equivalences: \arrow[ur, swap, equals, "1"] \end{tikzcd} \end{center} - for paths $p_l : x = y$, $p_b : y = z$, $q_l : h(x) = h(y)$, $q_b : h(y) = h(z)$ and for 2-paths $s_l : \mapfunc{h}(p_l) = q_l$ (for the top) and $s_b : \mapfunc{h}(p_b) = q_b$ (for the back). + for paths $p_l : x = y$, $p_b : y = z$, $q_l : h(x) = h(y)$, $q_b : h(y) = h(z)$ and for 2-paths $s_l : \mapfunc{h}(p_l) = q_l$ (for the top) and $s_b : \mapfunc{h}(p_b) = q_b$ (for the back). Again, a filler is provided by path induction on the squares and paths defining the cube. - \textbf{Case $f_2\judgeq 0$ (diagram on the right)}. (TODO) + (TODO: this homotopy is pointed) - To show that this homotopy is pointed, (TODO) + \textbf{Case $f_2\judgeq 0$}. The case for the diagram on the right in (\ref{eq:pent-statement}) is easier and can be resolved analogously. Again, we first apply induction on the paths that $f_1$, $g_1$ and $g_2$ respect the basepoint, reducing the diagram to the following square: + \begin{equation}\label{eq:pent-right-all} + \begin{tikzcd} + (0 \o f_1)\smsh (g_2 \o g_1) + \arrow[r, equals,"i"] + \arrow[d, swap, equals,"1"] + & (0 \smsh g_2)\o (f_1 \smsh g_1) + \arrow[d,equals,"t_{g_1}\o (f_2\smsh g_2)"] + \\ + 0 \smsh (g_2 \o g_1) + \arrow[r, swap, equals,"t_{g_1 \o g_2}"] + & 0 + \end{tikzcd} + \end{equation} + For the underlying homotopy. let $x : A_1 \smsh B_1$ and apply induction on $x$. For the point constructors: if $x \judgeq (a,b)$, $x \judgeq \auxl$ or $x \judgeq \auxr$, we have to fill (respectively) the squares: + \begin{equation}\label{eq:pent-right-ab} + \begin{tikzcd}[column sep=5em] + (a_3,g_2(g_1(b))) + \arrow[r, equals,"1"] + \arrow[d,swap,equals,"1"] + %\arrow[d,equals,"\gluer_{g_2(g_1(b))}\tr\gluer_{g_2(g_1(b_1))}\sy"] + & (a_3,g_2(g_1(b))) + \arrow[d,equals,"\gluer_{g_2(g_1(b))}\tr\gluer_{b_3}\sy"] + \\ + (a_3,g_2(g_1(b))) + \arrow[r,swap,equals,"\gluer_{g_2(g_1(b))}\tr\gluer_{b_3}\sy"] + & (a_3,b_3) + \end{tikzcd} + \end{equation} + \begin{equation}\label{eq:pent-right-aux} + \begin{tikzcd}[column sep=5em] + \auxl + \arrow[r, equals,"1"] + \arrow[d,swap,equals,"1"] + & \auxl + \arrow[d,equals,"\gluel_{a_3}\sy"] + & \auxr + \arrow[r, equals, "1"] + \arrow[d, swap, equals, "1"] + & \auxr + \arrow[d, equals, "\gluer_{b_3}\sy"] + \\ + \auxl + \arrow[r,swap, equals,"\gluel_{a_3}\sy"] + & (a_3, b_3) + & \auxr + \arrow[r, swap, equals, "\gluer_{b_3}\sy"] + & (a_3, b_3) + \end{tikzcd} + \end{equation} + which can be done by reflexivity. The cubes to be filled for $x$ varying over $\gluel_a$ or $ gluel_b$ are similar to the ones in (\ref{eq:pent-left-gluel}) and (\ref{eq:pent-left-gluel}) and can be generalized accordingly. (ADD REMARK cubically, ap id p reduces to refl, so the cases are exactly the same). + (TODO: this homotopy is pointed) \end{proof} +\begin{lem} + (TODO: the fillers in \autoref{lem:smash-coh} are the same) +\end{lem} + \begin{thm}\label{thm:smash-functor-right} Given pointed types $A$, $B$ and $C$, the functorial action of the smash product induces a map $$({-})\smsh C:(A\pmap B)\pmap(A\smsh C\pmap B\smsh C)$$