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\section*{experimental alternative definition of functor}

Good news! The twists and turns on this page have culminated in \href{http://ncatlab.org/nlab/show/functor#OnGeneralizedElements}{Functors and generalized elements}. From there, I drew the diagram



Staring at this diagram resulted in the following:


\vspace{.5em} \hrule \vspace{.5em}
\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

Given categories $C$ and $D$, a \textbf{functor} $F:C\to D$ is a map that sends each object $x$ in $C$ to an object $F(x)$ in $D$ and each morphism $f:x\to y$ in $C$ to a morphism $F(f):F(x)\to F(y)$ in $D$ such that each commuting diagram in $C$ maps to a commuting diagram in $D$.


\vspace{.5em} \hrule \vspace{.5em}
This has now been included in the definition of [[functor]].

Note this is still essentially the same thing as conveyed in my earlier picture



requiring the square commutes, but is done in a way that avoids introducing the non-standard morphism-like components $\alpha_x:x\to F(x)$.

In other words, if we stopped worrying about what $\alpha_x:x\to F(x)$ really was and simply accepted that the indicated square commutes, it implies automatically that any commuting diagram in $A$ maps to a commuting diagram in $B$.

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

This page is an informal/speculative discussion of an alternative (yet hopefully equivalent) definition of [[functor]]. It first appeared as a discussion at [[functor]] itself, but was subsequently moved here.

\noindent\hyperlink{discussion_30}{Discussion 3.0}\dotfill \pageref*{discussion_30} \linebreak
\noindent\hyperlink{discussion_20}{Discussion 2.0}\dotfill \pageref*{discussion_20} \linebreak
\noindent\hyperlink{discussion_10}{Discussion 1.0}\dotfill \pageref*{discussion_10} \linebreak
\noindent\hyperlink{definition_3}{Definition}\dotfill \pageref*{definition_3} \linebreak
\noindent\hyperlink{definition_4}{Definition}\dotfill \pageref*{definition_4} \linebreak
\noindent\hyperlink{definition_5}{Definition}\dotfill \pageref*{definition_5} \linebreak
\noindent\hyperlink{discussion}{Discussion}\dotfill \pageref*{discussion} \linebreak


\hypertarget{discussion_30}{}\subsection*{{Discussion 3.0}}\label{discussion_30}

I think I may have found a way to express this idea without resorting to non-standard morphism-like maps $x\to F(x)$ or cographs.

\hypertarget{definition_2}{}\subsection*{{Definition}}\label{definition_2}

Given categories $A$ and $B$, a map $F:A\to B$ is a functor if for every morphism $f:x\to y$ in $A$

\begin{displaymath}
F(f) F(x) = F(f(x)).
\end{displaymath}
Let's check\ldots{}

\begin{displaymath}
\itexarray{
F(g\circ f) F(x) 
&= F(g\circ f(x)) \\
&= F(g)\circ F(f(x)) \\
&= F(g)\circ F(f) F(x).}
\end{displaymath}
Since this is true for all morphisms $f:x\to y$ and $g:y\to z$, we have

\begin{displaymath}
F(g)\circ F(f) = F(g\circ f)
\end{displaymath}
as required.

Next,

\begin{displaymath}
F(1_A) F(x) = F(1_A(x)) = F(x)
\end{displaymath}
as required.

Therefore, it seems to work.

Note this is still essentially the same thing as conveyed in my earlier picture



requiring the square commutes, but is done in a way that avoids introducing the non-standard morphism-like components.

\hypertarget{discussion_20}{}\subsection*{{Discussion 2.0}}\label{discussion_20}

After Discussion 1.0 below, it seems that the original intention of this page was to define a [[functor]] in terms of a [[cograph of a functor|cograph]]. Given a functor, it is straightforward to construct its cograph. However, it may be possible to do the reverse, i.e. start with a cograph and infer the functor that gives rise to that cograph.

Is it possible to define a functor in terms of cograph? If so, how?

\hypertarget{discussion_10}{}\subsection*{{Discussion 1.0}}\label{discussion_10}

\hypertarget{definition_3}{}\subsubsection*{{Definition}}\label{definition_3}

Given small categories $A$ and $B$ define their \textbf{disjoint union} $A\sqcup B$ to be the category with

\begin{displaymath}
Obj(A\sqcup B) = Obj(A)\sqcup Obj(B)
\end{displaymath}
and

\begin{displaymath}
Hom_{A\sqcup B}(x,y) = 
  \left\{
    \begin{aligned}
      Hom_A(x,y) & if x,y \in A \\
      Hom_B(x,y) & if x,y \in B \\
      \emptyset & otherwise
    \end{aligned}
  \right.
\end{displaymath}
There are two \textbf{inclusion maps}

\begin{displaymath}
i_A:A\to A\sqcup B\quad\text{and}\quad i_B:B\to A\sqcup B
\end{displaymath}
defined for any morphism $f:a\to b$ in $A$ and morphism $g:c\to d$ in $B$ by

\begin{displaymath}
i_A(a) = a\sqcup\emptyset,\quad i_A(b) = b\sqcup\emptyset,\quad\text{and}\quad i_A(f) = f\sqcup\emptyset
\end{displaymath}
and

\begin{displaymath}
i_B(c) = \emptyset\sqcup\c,\quad i_B(d) = \emptyset\sqcup d,\quad\text{and}\quad i_B(g) = \emptyset\sqcup g.
\end{displaymath}
\hypertarget{definition_4}{}\subsubsection*{{Definition}}\label{definition_4}

Given categories $A$, $B$ and inclusion maps $i_A:A\to A\sqcup B$, $i_B:B\to A\sqcup B$, a \textbf{functor} is a map $F:A\to B$ together with component morphisms $\alpha_x:i_A(x)\to i_B\circ F(x)$ and $\alpha_y:i_A(y)\to i_B\circ F(y)$ for any morphism $f:x\to y$ in $A$ such that the following diagram commutes:

\begin{displaymath}
\itexarray{ 
    i_A(x)
    & 
    \stackrel{i_A(f)}{\to} 
    & 
    i_A(y) 
    \\ 
    \mathllap{\scriptsize{\alpha_x}}\downarrow 
    && 
    \downarrow\mathrlap{\scriptsize{\alpha_y}} 
    \\ i_B\circ F(x)
    & 
    \stackrel{i_B\circ F(f)}{\to} & i_B\circ F(y) 
  }
\end{displaymath}



\vspace{.5em} \hrule \vspace{.5em}
[[nlab:Todd Trimble|Todd]]: If $A \sqcup B$ means the disjoint union of $A$ and $B$, then there are no morphisms of the form $i_A(x) \to i_B(F(x))$.

[[Eric]]: Hi Todd. I saw your comment after I added the figure above. I am probably not expressing myself clearly. I want to get $Obj(A)$ and $Obj(B)$ into the same category while preserving all morphisms and then ADD components $\alpha_x$ and $\alpha_y$. Is that kind of thing allowed?

I think the picture expresses what I'm trying to do, but maybe I'm converting that into the wrong formulas.

[[Eric]]: The idea is that by requiring every such square to commute, you automatically get $F(g\circ f) = F(g)\circ F(f)$ and $F(Id_x) = Id_{F(x)}$. Or so I think\ldots{}

[[Eric]]: Not to mention, this \emph{looks} more like a natural transformation, so a pattern is more apparent.

[[Todd Trimble|Todd]]: What I was saying is that wherever these $\alpha_x$ are supposed to live, it's can't be in $A \sqcup B$, because in $A \sqcup B$, there simply are no morphisms of that form.

Your picture suggests that you have in mind some category $C$ into which $A$ and $B$ embed; it can't be $A \sqcup B$ since your $\alpha_x$ simply don't exist there. But it's some $C$ in which such morphisms $\alpha_x$ exist, let's say. Then it looks like your notion of functor involves the following things:

\begin{itemize}%
\item Embeddings (full embeddings?) $i_A: A \to C$, $i_B: B \to C$ into some category $C$.


\item A rule which assigns to each object $x$ of $A$ an object $F(x)$ of $B$, and to each morphism $f: x \to y$ of $A$ a morphism $F(f): F(x) \to F(y)$ of $B$. In other words, a map $F: A \to B$ between the underlying directed graphs of $A$ and $B$.


\item A rule which assigns to each object $x$ of $A$ a morphism $\alpha_x: i_A(x) \to i_B(F(x)$, making that square commute.



\end{itemize}
Okay, so far these conditions don't ensure functoriality: that $F(g \circ f) = F(g) \circ F(f)$ and $F(1_x) = 1_{F(x)}$. You're going to have to add more conditions to make that inference. For example, suppose $C$ has just one morphism from each object $i_A(a)$ to each object of $i_B(b)$. Then commutativity of those squares is automatic, for any directed graph morphism $F: A \to B$. Insofar as not all graph morphisms are functorial, you can't infer functoriality.

You could ask that all the $\alpha_x$ be isomorphisms. Then commutativity plus that condition would give functoriality, and your concept amounts to that of ordinary functor $F$ together with an ordinary natural isomorphism $i_A \overset{\sim}{\to} i_B \circ F$.

But the trouble with all this is that I have no idea what $C$ is supposed to be! It can't be $A \sqcup B$. Would it be something constructed in terms of $A$ and $B$ (and if so, what)? If it's just \emph{any} old category into which $A$ and $B$ embed and for which such isos $\alpha_x$ exist (and compatible with the $F(f)$ via commutative diagrams), then I have problems with that too.

\begin{description}

\item[[[Eric]]; Hi Todd. I think I'm confused because with the standard definition of functor $F:A\to B$ no one ever complains about writing $F(x)\in B$, so in a way, isn't $F$ an \emph{arrow} from $x\in A$ to $F(x)\in B$? We also write $F(f)\in B$, which makes me think $F$ is a bunch of 1-arrows and 2-arrows. I mean, doesn't the picture make sense? Why is it so hard to convert the picture to a mathematical statement?] \emph{Toby}: Eric, here you seem to be conflating these two arrows: $\to$ and $\mapsto$. At the level of functions between sets, if $A$ and $B$ are sets and $f$ is a function from $A$ to $B$, then I write $f\colon A \to B$ or $A \overset{f}\to B$. Then if $x$ is an element of $A$ and $y$ is an (the) element of $B$ such that $y = f(x)$, then I write $f\colon x \mapsto y$ or $x \overset{f}\mapsto y$. Or without bothering with $y$, I write $f\colon x \mapsto f(x)$ or $x \overset{f}\mapsto f(x)$. (Actually, this is best when `$x$' is a variable and `$f(x)$' is a formula in terms of that variable; for example, if $f$ is the squaring function, then I write $f\colon x \mapsto x^2$ or $x \overset{f}\mapsto x^2$, and this serves to \emph{define} $f$.)

 
Actually, you can find a lot of literature where the same symbol `$\rightarrow$' is used for both $\to$ and $\mapsto$, but they are still different concepts.

 
Of course, what the [[cograph]] gets you is that it turns $\mapsto$ into a special case of $\to$. But they are still different; $f\colon A \to B$ is an arrow in the category of sets or (in the categorified case) in the $2$-category of categories, while $x \overset{f}\mapsto y$ is (only in the categorified case) an arrow in the cograph of $f$.

 
\end{description}
\begin{quote}%
\begin{quote}%
[[Eric]]: Hmm.. but we ARE talking about cographs :) At least I was accidentally :)


\end{quote}

\end{quote}
[[Urs Schreiber]]: concerning the nature of $C$: as I suggested before, it does make sense to take this to be the [[cograph of a functor]]. And indeed, in some situations it is useful to \emph{define} the notion of functor in terms of the notion of cograph. In [[Higher Topos Theory]] the notion of [[adjoint (infinity,1)-functor]] is defined entirely in terms of cographs of functors.

[[Eric]]: Thanks Urs. Now I just need to parse what you say into something I can understand, but it is encouraging. In fact, quoting myself from earlier in the discussion:

\begin{quote}%
It seems like maybe I am trying to work backwards, i.e. define a functor via a cograph.


\end{quote}
When I tried to grok ``cograph'' on my personal web [[ericforgy:Natural Transformation]], I stripped out the line including the functor and with Toby's help, realized what I (think I) was talking about was the disjoint union of categories.

I guess, my plan of attack now is to see how we can define functor via cograph. For some reason, that makes sense to me.

[[Todd Trimble|Todd]]: Yes, that's true, the cograph is fine. I thought that Eric had wanted a $C$ which would uniformly work, independently of which functor $F$ was being considered. Instead, we construct the cograph $C$ in terms of $F$, and everything is for the best in this best of all possible worlds. And we get the picture Eric was aiming for.

[[Eric]]: Neat. Now, if we (I'll try, but I'm not confident and wouldn't mind if someone else did it!) can write down a succinct definition of functor in terms of cographs with a picture like the one above (if not that one), we could add it to [[functor]].

[[Eric]]: By the way, here is something I wrote on my personal web [[ericforgy:Natural Transformation]]:

\begin{quote}%
[[Urs Schreiber]]: at [[cograph]] it is indicated how that concept encodes the notion of [[profunctor]] which is a bit more general than that of [[functor]].

[[Eric]]: Oh! That is pretty! Thanks. I like this diagram

\begin{displaymath}
\itexarray{
\end{displaymath}

\end{quote}
\begin{verbatim}R &\to& {*}
\\
\downarrow && \downarrow
\\
X \times Y &\stackrel{\chi_R}{\to}& \{0,1\}\end{verbatim}
\}

\begin{displaymath}
>I need to think about it, but a vale of fog has been lifted. More fog remains, but thanks!

***

###Definition###
Given small categories $A$ and $B$ define their **disjoint union** $A\sqcup B$ to be the category with
$$Obj(A\sqcup B) = Obj(A)\sqcup Obj(B)
\end{displaymath}
and

\begin{displaymath}
Hom_{A\sqcup B}(x,y) = 
  \left\{
    \begin{aligned}
      Hom_A(x,y) & if x,y \in A \\
      Hom_B(x,y) & if x,y \in B \\
      \emptyset & otherwise
    \end{aligned}
  \right.
\end{displaymath}
There are two \textbf{inclusion maps}

\begin{displaymath}
i_A:A\to A\sqcup B\quad\text{and}\quad i_B:B\to A\sqcup B
\end{displaymath}
defined for any morphism $f:a\to b$ in $A$ and morphism $g:c\to d$ in $B$ by

\begin{displaymath}
i_A(a) = a\sqcup\emptyset,\quad i_A(b) = b\sqcup\emptyset,\quad\text{and}\quad i_A(f) = f\sqcup\emptyset
\end{displaymath}
and

\begin{displaymath}
i_B(c) = \emptyset\sqcup\c,\quad i_B(d) = \emptyset\sqcup d,\quad\text{and}\quad i_B(g) = \emptyset\sqcup g.
\end{displaymath}

\vspace{.5em} \hrule \vspace{.5em}
I intentionally did not use the word ``functor'' above even though that is really the [[coproduct]] in [[Cat]]. The reason is that I want to use this disjoint union to \emph{define} a functor. The following is a possible starting point:

\hypertarget{definition_5}{}\subsubsection*{{Definition}}\label{definition_5}

Given categories $A$, $B$ and inclusion maps $i_A:A\to A\sqcup B$, $i_B:B\to A\sqcup B$, a \textbf{functor} is a map $F:A\to B$ that assigns morphisms $\alpha_x:i_A(x)\to i_B\circ F(x)$ and $\alpha_y:i_A(y)\to i_B\circ F(y)$ for any morphism $f:x\to y$ in $A$ such that the following diagram commutes:

\begin{displaymath}
\itexarray{ 
    i_A(x)
    & 
    \stackrel{i_A(f)}{\to} 
    & 
    i_A(y) 
    \\ 
    \mathllap{\scriptsize{\alpha_x}}\downarrow 
    && 
    \downarrow\mathrlap{\scriptsize{\alpha_y}} 
    \\ i_B\circ F(x)
    & 
    \stackrel{i_B\circ F(f)}{\to} & i_B\circ F(y) 
  }
\end{displaymath}



\vspace{.5em} \hrule \vspace{.5em}
\hypertarget{discussion}{}\subsubsection*{{Discussion}}\label{discussion}

[[Eric]]: Motivated by some discussion over at [[natural transformation]], I was wondering if the following alternative definition of functor holds water:

\begin{quote}%
\hypertarget{definition_6}{}\subsubsection*{{Definition}}\label{definition_6}

Given categories $A$, $B$, a \textbf{functor} is a map $F:A\to B$ that assigns maps $\alpha_x:x\to F(x)$ and $\alpha_y:y\to F(y)$ for any morphism $f:x\to y$ in $A$ such that the following diagram commutes:

\begin{displaymath}
\itexarray{
\end{displaymath}

\end{quote}
\begin{verbatim}x
& 
\stackrel{f}{\to} 
& 
y 
\\ 
\mathllap{\scriptsize{\alpha_x}}\downarrow 
&& 
\downarrow\mathrlap{\scriptsize{\alpha_y}} 
\\ F(x) 
& 
\stackrel{F(f)}{\to} & F(y) \end{verbatim}
\}

\begin{displaymath}
If so, it has a certain beauty that appeals to me. Thanks!

Urs: that diagram seems to be meaningful only if $A = B$, right? In that case, this is a functor $F : A \to A$ together with a natural transformation from the identity functor on $A$ into $F$.

[[Eric]]: Yeah. It needs some work. Of course I don't want $A=B$, but I would like to be able to introduce morphisms from $A$ to $B$. Could I form some kind of category union or something? The picture is kind of neat. Each diagram in $A$ "hovers" above a diagram in $B$. If it can be ironed out, it might be an efficient definition.

Urs: now you are looking for the notion [[cograph of a functor]] (you should make us write out more details at that entry!)

[[Eric]]: **Gulp!** Ok. But is there a term for the (small) category $C = A\cup B$ where $C_0 = A_0\cup B_0$ and $C_1 = A_1\cup B_1$?

_Toby_:  You could just call that the _union_ of the two categories.  But it\'s not really a meaningful concept unless you already have $A$ and $B$ as [[subcategory|subcategories]] of some ambient category $D$, or something like that.  To ask whether an element of $A_0$ matches an element an element of $B_0$ (and similarly for $A_1$ and $B_1$) is [[evil]].

Alternatively, you could take it as understood that the sets $A_0$ and $B_0$ are disjoint, (and similarly for $A_1$ and $B_1$).  Then you get the _disjiont union_ of the two categories, and that is an important concept; it is the [[coproduct]] in [[Cat]].  And that is one of the ingredients of the [[cograph of a functor]], so probably it\'s what you want.

[[Eric]]: It seems like maybe I am trying to work backwards, i.e. define a functor via a cograph.

[[Eric]]: Ok! I think I got it. Does this hold water?

***

###Definition###
Given categories $A$, $B$ and inclusion maps $i_A:A\to A\sqcup B$, $i_B:B\to A\sqcup B$, a **functor** is a map $F:A\to B$ that assigns morphisms $\alpha_x:i_A(x)\to i_B\circ F(x)$ and $\alpha_y:i_A(y)\to i_B\circ F(y)$ for any morphism $f:x\to y$ in $A$ such that the following diagram commutes:
\end{displaymath}
$\backslash$array\{ i\_A(x) \& $\backslash$stackrel\{i\_A(f)\}\{$\backslash$to\} \& i\_A(y) $\backslash$ $\backslash$mathllap\{$\backslash$scriptsize\{$\backslash$alpha\_x\}\}$\backslash$downarrow \&\& $\backslash$downarrow$\backslash$mathrlap\{$\backslash$scriptsize\{$\backslash$alpha\_y\}\} $\backslash$ i\_B$\backslash$circ F(x) \& $\backslash$stackrel\{i\_B$\backslash$circ F(f)\}\{$\backslash$to\} \& i\_B$\backslash$circ F(y) \}

\begin{displaymath}
It follows that $i_A$ and $i_B$ are functors.

Note that a map $F:A\to B$ is a functor if the map $\alpha:i_A\Rightarrow i_B\circ F$ is a natural transformation.

***

Urs: to be frank, I don't get it! Also: what is it you are really after? The concept of a functor is entirely non-mysterious, it seems. What are you dissatisfied with, with the standard definition?

[[Eric]]: I enjoyed fiddling with [[natural transformations]] recently. I hadn't thought about them much. The introduction of "components" was kind of interesting. I also liked that "sheet transformation", which I now think of as a some kind of 3-morphism. So I was wondering if we could introduce "components" to define functors (which I now think of as 2-morphisms). Then I wondered if we could continue a pattern downward all the way to (-2)-categories.

The _pattern_ natural transformation $\to$ functor $\to$ morphism is not 100% clear to me. Each individually "makes sense", but I'd like to see a clearer pattern how they're related.

Basically, I'm trying to understand

* (-2)-morphisms
* (-1)-morphisms
* 0-morphisms
* 1-morphisms (morphisms)
* 2-morphisms (functors?)
* 3-morphisms (natural transformations? sheet transformations?)
* Etc.

in some consistent way.

For example, with categories $A$ and $B$ we have a [[functor category]] $[A,B]$ which is related to (if not equal to) a 2-category. How about sets $A$ and $B$ thought of as 0-categories? We could form a "morphism category" $[A,B]$ that is related (if not equal to) a 1-category. Then I can imagine a "truth category" $[True,True]$ that is related (if not equal) to a 0-category. Etc etc. This is all pretty elementary stuff I'm sure.

In a nutshell, I'm not saying there is anything wrong the definitions there, but I'm trying to understand them in a different way if possible.

##Reference

* See all [[experimental alternative definition of adjunction]].

  

[[!redirects functor (discussion)]]
\end{displaymath}


\end{document}