**Warning: These are my notes trying to understand natural transformations. Feedback welcome!**

The thing I’m hoping to understand here is:

Why are natural transformation defined for functors sharing the same domain and codomain?

Eric: For some reason, I always thought a natural transformation was a more general map between functors $\alpha:F\Rightarrow G$, where

$F:A\to B\quad\text{and}\quad G:C\to D.$

Then $\alpha(F):C\to D$ is a functor and we’d probably also want 2-component functors $\alpha_C:A\to C$ and $\alpha_D:B\to D$ such that the following diagram commutes

$\array{
A
&
\stackrel{F}{\to}
&
B
\\
\mathllap{\scriptsize{\alpha_C}}\downarrow
&\mathllap{\scriptsize{\alpha}}\Downarrow &
\downarrow\mathrlap{\scriptsize{\alpha_D}}
\\ C
&
\stackrel{G}{\to} & D
}$

If this is not a natural transformation, is there another name for it?

Toby: What is that $\mathllap{\scriptsize{\alpha}}\Downarrow$ in the middle? The only sense that I can make of it is that it is itself a natural transformation (in the standard sense) between the functors $F; \alpha_D$ and $\alpha_C; G$. In that case, I would call this thing a

lax commutative square(although possibly it iscolaxinstead). If you have $\mathllap{\scriptsize{\alpha}}\Uparrow$ instead, that would be acolax commutative square(unless I have them backwards); if you specify that $\mathllap{\scriptsize{\alpha}}\Downarrow$ is a naturalisomorphism, then you get aweak commutative square(the default notion ofcommutative squarein my mind); if you are working with strict categories (with an inherent notion of equality of objects) and functors, then you can even demand that $\mathllap{\scriptsize{\alpha}}\Downarrow$ is an identity, and then you get astrict commutative square. These are all notions of morphism (butnot$2$-morphism) in an appropriate arrow $2$-category.Compare the arrow category of, say, the category of sets. If $A,B,C,D$ are sets and $F,G$ are functions, then $F$ and $G$ are objects in the arrow category $Set^\mathbf{2}$, and a morphism $\alpha\colon F \to G$ in this arrow category is a commutative square (there is only one sense here)

`\array{`

```
A
&
\stackrel{F}{\to}
&
B
\\
\mathllap{\scriptsize{\alpha_C}}\downarrow
&\mathllap{\scriptsize{=}}\Downarrow &
\downarrow\mathrlap{\scriptsize{\alpha_D}}
\\ C
&
\stackrel{G}{\to} & D
```

}

$---Toby
a class=existingWikiWord href=/ericforgy/published/EricEric/a: Thanks Toby. The first time around, I misread your comment (probably because of a parallel discussion going on elsewhere). This a class=existingWikiWord href=/nlab/show/arrow+categoryarrow category/a thing looks like what Im after.
a class=existingWikiWord href=/ericforgy/published/EricEric/a: When$

\array{ A & \stackrel{F}{\to} & B \ \mathllap{\scriptsize{\alpha_C}}\downarrow &\mathllap{\scriptsize{\alpha}}\Downarrow & \downarrow\mathrlap{\scriptsize{\alpha_D}} \ C & \stackrel{G}{\to} & D }

$to commute, we get neat (and familiar) component diagrams$

\array{ \alpha_D\circ F(x) & \stackrel{\alpha_D\circ F(f)}{\to} & \alpha_D\circ F(y) \ \downarrow && \downarrow \ G\circ\alpha_C(x) & \stackrel{G\circ\alpha_C(f)}{\to} & G\circ\alpha_C(y) } \quad\quad\text{and}\quad\quad \array{ G\circ\alpha_C(x) & \stackrel{G\circ\alpha_C(f)}{\to} & G\circ\alpha_C(y) \ \downarrow && \downarrow \

\alpha_D\circ F(x) & \stackrel{\alpha_D\circ F(f)}{\to} & \alpha_D\circ F(y) }

```
This says that we have a natural transformation (in the standard) sense $\alpha_C\circ G\Rightarrow F\circ\alpha_D$.
**LIGHT BULB**
We don't want the diagram to commute. Sorry! If it commutes, it is not very interesting. It should commute "up to" a standard natural transformation $\alpha_C\circ G\Rightarrow F\circ\alpha_D$.
**Thinking...**
<a class='existingWikiWord' href='/ericforgy/published/Eric'>Eric</a>: I think I got it.
Given two functors $F:A\to B$ and $G:C\to D$, let $\alpha:F\Rightarrow G$ together with component functors $\alpha_C:A\to C$ and $\alpha_D:B\to D$ be a map such that
$$\alpha_D\circ F\Rightarrow G\circ\alpha_C
```

is a natural transformation (in the standard sense). Does this guy have a name?

The reason why I bring this up is that I do not like bigons. Bigons are not a good shape for doing computational geometry, geometric realization, etc. So before I could finally grok what a cone was, I had to draw a bunch of diagrams. These diagrams were bigons. Yucky!

Thanks Todd. That was a valiant effort, but didn’t manage to convince me (I have a thick skull!). I still like this more general definition. Given topological spaces $This says that we have a natural transformation (in the standard) sense$, we could also be interested in studying spaces $(W\to X)\Rightarrow (Y\to Z)$. What you described would then be a special case where $W=Y$ and $X=Z$.

Here is my attempt to formalize an alternative definition:

## Definition

Given categories $A$,$B$,$C$,$D$ and functors $F:A\to B$,$G:C\to D$ a

natural transformation$\alpha:F \Rightarrow G$`\array{`

```
A
&
\stackrel{F}{\to}
&
B
\\
\mathllap{\scriptsize{\alpha_C}}\downarrow
&\mathllap{\scriptsize{\alpha}}\Downarrow &
\downarrow\mathrlap{\scriptsize{\alpha_D}}
\\ C
&
\stackrel{G}{\to} & D
```

}

$>assigns functors$

\array{ \alpha_D\circ F(x) & \stackrel{\alpha_D\circ F(f)}{\to} & \alpha_D\circ F(y) \ \mathllap{\scriptsize{\alpha_x}}\downarrow && \downarrow\mathrlap{\scriptsize{\alpha_y}} \ G\circ\alpha_C(x) & \stackrel{G\circ\alpha_C(f)}{\to} & G\circ\alpha_C(y) }

```
***
Again, this reduces to the traditional definition when $\alpha_C$ and $\alpha_D$ are identity functors. What do you think? If this doesn't already have a name (it probably does), what would be a good name for it?
***
<a class='existingWikiWord' href='/ericforgy/published/Eric'>Eric</a>: I wrote the definition down intentionally so that the composite _is_ a natural transformation, but I hope I can convince you that $\alpha:F\Rightarrow G$ is _not_ a natural transformation because the domains and codomains of $F$ and $G$ do not coincide (which they must to be a standard natural transformation). I'll call it something else then. How about "sheet transformations"?
>##Definition##
>Given categories $A$,$B$,$C$,$D$ and functors $F:A\to B$,$G:C\to D$ a __sheet transformation__ $\alpha:F \Rightarrow G$
```

\array{ A & \stackrel{F}{\to} & B \ \mathllap{\scriptsize{\alpha_C}}\downarrow &\mathllap{\scriptsize{\alpha}}\Downarrow & \downarrow\mathrlap{\scriptsize{\alpha_D}} \ C & \stackrel{G}{\to} & D }

```
>assigns functors $\alpha_C:A\to C$ and $\alpha_D:B\to D$ called **component functors** such that
$$\alpha_D\circ F\Rightarrow G\circ\alpha_C
```

is a

natural transformation.

Ok! I like that. Now let’s see if I can go down in dimensions.

**Under Construction**

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

## Definition

Given categories $*** Again, this reduces to the traditional definition when$, $B$, a

functoris 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:`\array{`

```
x
&
\stackrel{f}{\to}
&
y
\\
\mathllap{\scriptsize{\alpha_x}}\downarrow
&&
\downarrow\mathrlap{\scriptsize{\alpha_y}}
\\ F(x)
&
\stackrel{F(f)}{\to} & F(y)
```

}

```
If so, it has a certain beauty that appeals to me. 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)
```

and

$If so, it has a certain beauty that appeals to me. Thanks!
***
##Definition##
Given small categories$

There are two **inclusion maps**

$i_A:A\to A\sqcup B\quad\text{and}\quad i_B:B\to A\sqcup B$

defined for any morphism $f:a\to b$ in $A$ and morphism $g:c\to d$ in $B$ by

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

and

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

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:

$\array{
i_A(x)
&
\stackrel{i_A(f)}{\to}
&
i_A(y)
\\
\mathllap{\scriptsize{\alpha_x}}\downarrow
&&
\downarrow\mathrlap{\scriptsize{\alpha_y}}
\\ i_B\circ F(x)
&
\stackrel{i_B\circ F(f)}{\to} & i_B\circ F(y)
}$

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.

Compare the above definition to

Given categories $A$,$B$,$C$,$D$ and functors $F:A\to B$,$G:C\to D$ a **natural transformation** $\alpha:F \Rightarrow G$

$\array{
A
&
\stackrel{F}{\to}
&
B
\\
\mathllap{\scriptsize{\alpha_C}}\downarrow
&\mathllap{\scriptsize{\alpha}}\Downarrow &
\downarrow\mathrlap{\scriptsize{\alpha_D}}
\\ C
&
\stackrel{G}{\to} & D
}$

assigns functors $\alpha_C:A\to C$ and $\alpha_D:B\to D$ called **component functors** and for any morphism $f:x\to y$ in $A$ assigns morphisms $\alpha_x:\alpha_D\circ F(x)\to G\circ\alpha_C(x)$ and $\alpha_y:\alpha_D\circ F(y)\to G\circ\alpha_C(y)$ in $D$ called **component morphisms** such that the following diagram commutes in $D$:

$\array{
\alpha_D\circ F(x)
&
\stackrel{\alpha_D\circ F(f)}{\to}
&
\alpha_D\circ F(y)
\\
\mathllap{\scriptsize{\alpha_x}}\downarrow
&&
\downarrow\mathrlap{\scriptsize{\alpha_y}}
\\ G\circ\alpha_C(x)
&
\stackrel{G\circ\alpha_C(f)}{\to} & G\circ\alpha_C(y)
}$

A response to Urs at experimental alternative definition of functor

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.

I'm trying to understand why you think functors are $2$-morphisms! —Toby

Eric: Notice the question mark? :) I still haven’t SEEN a definition telling me what a 2-morphism is except in a globular 2-category (which I should review).

Toby: I noticed it, but maybe there was something behind it?Functors go

betweencategories or higher categories, while $2$-morphisms liewithina given higher category. Of course, you could have a higher category $C$ofcategories or higher categories, in which case functors would lie within $C$, and you can even set this up so that the $2$-morphisms in $C$ are functors (although it is more common to set things up so that the $1$-morphisms or $0$-morphisms are functors). However, that would be a specific example (or family of examples), not the general idea.In a category, as you know, the objects might be anything, and the morphisms might be anything. So also, in a higher category, the $2$-morphisms might be anything. If you look at a specific definition of some kind of higher category, it is likely that ‘$2$-morphism’ will be an undefined term, as it is in the definition of strict 2-category. It is possible that ‘$2$-morphism’ might not be one of the basic undefined terms but instead defined in terms of something else, but something else will still be an undefined term, and it is probably much more common to define things so that ‘$2$-morphism’

isa basic term. So I would write your list above as

- $0$-morphisms
- $1$-morphisms
- $2$-morphisms
- $3$-morphisms
- etc
Eric: Right. 2-morphisms can

beanything, but I’d think they have to satisfy some conditions in terms of objects and morphisms. But that seems to depend on theshapesused to define your 2-category. Bigons are one shape and there you get things like horizontal and vertical composition and interchange laws, but what about for “squares” or some other shape?with nothing in parentheses. (While most people would start at $0$-morphisms, I am willing to do the negative thinking that considers $k$-morphisms for $k \lt 0$. As far as I can see, however, this wouldn't be enlightening for you.)

A very important example $C$ is the $2$-category Cat of categories, in which we have

- $0$-morphisms (categories)
- $1$-morphisms (functors)
- $2$-morphisms (natural transformations)
- $3$-morphisms (equations)
- $k$-morphisms for $k \gt 3$ (trivial).
More generally, we have in the $(n+1)$-category $n Cat$ of $n$-categories

- $0$-morphisms ($n$-categories)
- $1$-morphisms (functors)
- $2$-morphisms (natural transformations)
- $3$-morphisms (modifications)
- etc.
- $(n+1)$-morphisms (equations)
- $k$-morphism for $k \gt n + 1$ (trivial).
And in the $\infty$-category $\infty Cat$ of $\infty$-categories, we have

- $0$-morphisms ($\infty$-categories)
- $1$-morphisms (functors)
- $2$-morphisms (natural transformations)
- $3$-morphisms (modifications)
- etc (never trivial).
For an example of a different sort, let $A$ and $B$ be specific categories, and let $C$ be the functor category of all functors from $A$ to $B$. Then we have

- $0$-morphisms (functors from $A$ to $B$)
- $1$-morphisms (natural transformations)
- $2$-morphisms (equations)
- $k$-morphisms for $k \gt 2$ (trivial).
Or go back to the $2$-category $Cat$ of categories, with the monoidal structure given by the cartesian product, and consider its delooping $\mathbf{B} Cat$. Then we have

- $0$-morphisms (trivial)
- $1$-morphisms (categories)
- $2$-morphisms (functors)
- $3$-morphisms (natural transformations)
- $4$-morphisms (equations)
- $k$-morphisms for $k \gt 4$ (trivial),
which I mention only because it's an example where $2$-morphisms are functors; I don't think that it's a particularly illuminating example.

All of these examples (even the last one, but less so) are important examples, which it's worth understanding some day. But I'm not sure how much help they'll be in understanding higher categories

in general, since what is true of these examples is not going to be true of most higher categories. And thinking about higher categories whose objects or morphisms are themselves categories or higher categories invites the confusion of level slips, so it may be best toavoidthese examples until you already have a clear idea of the basic concepts.Eric: Thanks Toby! That is interesting. And yes I am interested in the negative categories :) I’m trying to see a pattern. I think the key concept I need to understand is profunctor. I’ll need some time to absorb what you said here.

Toby: If you find it easiest to understand functions as certain relations, then presumably you'll find it easiest to understand functors as certain profunctors. (Personally, I'm the opposite, but not everybody is.)

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.

Next, I want to continue the pattern and define morphisms this way!

**Under Construction**

Let $A$ be a category with objects $x,y$ with a inclusion maps $i_x:x\to A$ and $i_y:y\to A$.

A map $f:x\to y$ is a morphism if $i_A(f):i_A(x)\to i_A(y)$ is a functor.

Revised on February 21, 2010 at 01:19:36
by
Eric Forgy