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\begin{document}

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\section*{experimental alternative definition of adjunction}

This page is a result of the following question originally asked at [[adjunction]]:

[[Eric]]: Is it true that given categories $C$ and $D$ that functors $F:C\to D$ and $G:D\to C$ form an adjunction if the following diagram commutes for all morphisms $f:x\to y$ in $C$?

\begin{displaymath}
\itexarray{
  x &\stackrel{f}{\to}& y
  \\
  \uparrow && \uparrow
  \\
  G\circ F(y) 
  &\stackrel{G\circ F(f)}{\leftarrow}&
  G\circ F(x)
 }
\end{displaymath}
If so, is that an ``if and only if'' so that you can define adjunctions this way? If so, this would, in my opinion, be the most digestible way to define adjunctions for ``scientists and engineers''.

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

This page is an informal/speculative discussion of an alternative (yet hopefully equivalent) definition of [[adjunction]].

\noindent\hyperlink{attempt_2}{Attempt \#2}\dotfill \pageref*{attempt_2} \linebreak
\noindent\hyperlink{definition_experimental}{Definition (Experimental)}\dotfill \pageref*{definition_experimental} \linebreak
\noindent\hyperlink{attempt_1}{Attempt \#1}\dotfill \pageref*{attempt_1} \linebreak
\noindent\hyperlink{definition_experimental_2}{Definition (Experimental)}\dotfill \pageref*{definition_experimental_2} \linebreak
\noindent\hyperlink{response_from_toby}{Response from Toby}\dotfill \pageref*{response_from_toby} \linebreak


\hypertarget{attempt_2}{}\subsection*{{Attempt \#2}}\label{attempt_2}

\hypertarget{definition_experimental}{}\subsubsection*{{Definition (Experimental)}}\label{definition_experimental}

\hypertarget{attempt_1}{}\subsection*{{Attempt \#1}}\label{attempt_1}

\hypertarget{definition_experimental_2}{}\subsubsection*{{Definition (Experimental)}}\label{definition_experimental_2}

\begin{quote}%
Note the \textbf{Crash!} below. This definition is not correct. I think something like this should work, but I need to think about it.


\end{quote}
Given [[categories]] $C$ and $D$, a pair of [[functors]] $F:C\to D$ and $G:D\to C$ form an \textbf{adjunction} if for every morphism $f:x\to y$ in $C$, there are component morphisms

\begin{displaymath}
\eta_{y x}: G\circ F(x)\to y
\end{displaymath}
and

\begin{displaymath}
\eta_{x y}: G\circ F(y)\to x
\end{displaymath}
such that the following diagram commutes

\begin{displaymath}
\itexarray{
  x &\stackrel{f}{\to}& y
  \\
  \uparrow && \uparrow
  \\
  G\circ F(y) 
  &\stackrel{G\circ F(f)}{\leftarrow}&
  G\circ F(x)
 }.
\end{displaymath}
Note that since $G\circ F:C\to C$ is an [[endofunctor]], there are component morphisms

\begin{displaymath}
\alpha_x: x\to G\circ F(x)
\end{displaymath}
and

\begin{displaymath}
\alpha_y: y\to G\circ F(y)
\end{displaymath}
such that the following diagram also commutes:

\begin{displaymath}
\itexarray{
  x &\stackrel{f}{\to}& y
  \\
  \downarrow && \downarrow
  \\
  G\circ F(x) 
  &\stackrel{G\circ F(f)}{\rightarrow}&
  G\circ F(y)
 }.
\end{displaymath}
Putting the two commuting squares together, we have a commuting square with commuting (upward) diagonals giving rise to additional commuting triangles.

These commuting diagrams give rise to the following relations:

\begin{itemize}%
\item $\eta_{y x}\circ \alpha_x = f$


\item $\alpha_y\circ\eta_{y x} = G\circ F(f)$


\item $[G\circ F(f)]\circ\alpha_x = \alpha_y\circ f$


\item $\eta_{x y}\circ [G\circ F(f)]\circ\alpha_x = 1_x$


\item $f\circ\eta_{x y}\circ\alpha_y = 1_y.$



\end{itemize}
Note: I'm not sure if these are independent or not (need to count how many independent equations we get!), but it is instructive to note:

\begin{displaymath}
\eta_{x y}\circ \alpha_y\circ\eta_{y x}\circ\alpha_x = 1_x
\end{displaymath}
and

\begin{displaymath}
\eta_{y x}\circ\alpha_x\circ\eta_{x y}\circ \alpha_y = 1_y
\end{displaymath}
which means

\begin{displaymath}
\eta_{y x}\circ\alpha_x = \left(\eta_{x y}\circ \alpha_y\right)^{-1}.
\end{displaymath}
\textbf{CRASH!}

I'll need to check, but unless I made some algebra mistake, this last statement might kill the whole idea. Since

\begin{displaymath}
f = \eta_{y x}\circ\alpha_x,
\end{displaymath}
then the above statement says $f$ has an inverse.

\hypertarget{response_from_toby}{}\subsubsection*{{Response from Toby}}\label{response_from_toby}

You need to say more than just that \emph{there are} morphisms that make these diagrams commute. Whether $F \vdash G$ is an adjunction is not actually a property but a structure; we need to say \emph{how} they are an adjunction. That is, you have to pick the family of morphism $G(F(x)) \to y$ in advance.

But then you can make it work, I believe \ldots{} although I'm pretty sure that you need more commutative diagrams than the one that you've drawn. That is, an \textbf{adjunction} $\eta$ from $F$ to $G$ consists of a family of maps $\eta_f\colon G(F(x)) \to y$, one for each morphism $f\colon x \to y$, such that

\begin{itemize}%
\item a bunch of diagrams commute.

\end{itemize}
I have to think about whether this works and what diagrams you need.

[[Eric]]: Thanks Toby. I would be surprised if it were this simple, but I drew some pictures based on the diagram in Goldblatt, and I think this works.

[[Eric]]: Note, I had some typos in the original version of the diagram, but it should be correct now. Or it least it now matches the diagram on my paper :)

\begin{quote}%
I'm pretty sure that you need more commutative diagrams than the one that you've drawn.


\end{quote}
We get another commuting square from the fact that $G\circ F$ is an endofunctor. I've now drawn that above. When you put the two together, it gives you a bunch of additional commuting triangles, but everything follows from the one first simple diagram.

[[Eric]]: Hi Toby (and anyone else who may stop by). Note the \textbf{Crash!} above. The definition cannot be correct as stated. I probably won't give up though. I think instead of saying the diagram commutes, I need to say something about ``commuting up to a 2-morphism'' or ``there is a 2-morphism such that'' blah blah.



\end{document}