Abstract Nonsense

Natural Transformations (Pt. I)

Point of Post: In this post we introduce the notion of natural transformations, natural equivalences, and prove some basic theorems concerning them. We also give a few examples of natural transformations between some specific functors, as well as cover the classic natural equivalences (e.g. the natural equivalence of the double dual functor and the identity functor on $\mathbf{FinVect}_k$).

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Motivation

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In this post we discuss a notion in category theory which will finally make rigorous such wildly tossed around statements such as “____ is  naturally/canonically isomorphic to ___”. That phrase, or some variant thereof, is heard so often in elementary mathematics (e.g. basic group theory, even in books as basic as Pinter or Fraleigh, as well in non-algebraic books such as Simmons) but is rarely (if ever!) actually given meaning in those settings. So, what precisely does this mean? Often times one hears explanations of ‘canonically’ as meaning “without having to choose”. For example, an isomorphism in linear algebra is ‘canonical’ if we can formulate it without having to fix a basis in the source or target space. But, of course this has no meaning when one is dealing with isomorphisms in, say, $\mathbf{Top}$, as there are no choices ever really made. So, what does it mean for two things to be ‘naturally’ isomorphic? We begin by phrasing the problem in more categorical language. Often times one speaks of ‘natural isomorphisms’ between certain constructions. For example, in $\mathbf{FinVect}_k$ (the category of finite dimensional vector spaces over some field $k$) it’s often stated that “every finite dimensional vector space $V$  is ‘naturally’ isomorphic to its double dual $V^{\ast\ast}$“. Really what is being said, in a sense, is that the construction mapping $V$ to itself and the construction mapping $V$ to $V^{\ast\ast}$ are “the same”. Or, putting things in phrasing we know, that there is some notion of equivalence between the identity functor and double dual functor on $\mathbf{FinVect}_k$. What is this equivalence? While there are many explanations of the intuition behind the kind of equivalence we’re interested, there is a particularly beautiful description which, at the risk of alienating some readers, I’d like to explain. In topology given two maps $f,g:X\to Y$ we call them homotopic if there exists a continuous path from $f$ to $g$. Said differently, if there exists a continuous map $F:X\times I\to Y$ (where $I=[0,1]$) such that $F(x,0)=f$ and $F(x,1)=g$. One thinks of maps as being homotopic if one can nicely slide one path onto the other path without hitting any snags/tearing. One can think of this process as be indexed by time, and at each second there is a curve, and as time progresses this goes from looking like $f$ to looking like $g$. Intuitively, natural transformations are the homotopies of category theory. Namely, a natural transformation from a functor $F$ to a functor $G$ is a ‘continuous’ (i.e. respects the category, doesn’t “tear” anything) deformation of the functor $F$ into the functor $G$. Of course, doesn’t “tear” anything and respects the category should be further explained. By this we shall see, as I’d like to defer to the discussion below, an associated shifting of the “image” of $F$ to the “image” of $G$, and by not “tearing” and “respecting the category” we mean that this diagram commutes–that we can figure out information in the image of $G$ by working through the image of $F$ and this transformation between them.

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Natural Transformations

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Let $\mathcal{C}$ and $\mathcal{D}$ be two given categories and assume that that we have functors $\mathcal{C}\overset{\displaystyle \overset{F}{\longrightarrow}}{\underset{G}{\longrightarrow}}\mathcal{D}$. A natural transformation between $F$ and $G$ is a set $\eta=\{\eta_X\}$ of morphisms $\eta_X:F(X)\to G(X)$ for each $X\in\text{obj}(\mathcal{C})$ such that the following square is commutative for all $X,Y$ and all morphisms $X\xrightarrow{f}Y$

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$\begin{matrix} F(X) & \overset{\eta_X}{\longrightarrow} & G(X)\\ _{F(f)}\big\downarrow & & \big\downarrow _{G(f)}\\ F(Y) & \overset{\eta_Y}{\longrightarrow} & G(Y)\end{matrix}$

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We call this the naturality condition. We notate the fact that $\eta$ is a natural transformation from $F$ to $G$ by $\eta:F\implies G$. A natural transformation $\eta:F\implies G$ is called a natural equivalence if $F(X)\xrightarrow{\eta_X}G(X)$ is an isomorphism for each $X\in\text{obj}(\mathcal{C})$. If $\eta:F\implies G$ is a natural equivalence, we are apt to notate this as $\eta:F\cong G$.

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We see then that, as stated in the motivation, the set $\eta$ of transformations creates a natural (no pun intended) “shifting” of the image of $F$ to the image of $G$ as the following diagram (copied [laboriously I might add] and badly from [1]) indicates:

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$\begin{matrix} F(x) & & & \overset{\eta_x}{\longrightarrow} & G(x) & & \\ & \searrow^{F(g)} & & & &\searrow^{G(g)} \\ _{F(f)}\big\downarrow& & F(y) & \overset{\eta_y}{\longrightarrow} & \big\downarrow_{G(f)} & & G(y)\\ & \nearrow_{F(h)} & & & & \nearrow_{G(h)}& \\ F(z)& & & \overset{\eta_z}{\longrightarrow} & G(z) & & \\ & & & & & \end{matrix}$

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Moreover, the naturality conditions is precisely the statement that this diagram commutes! Thus, the idea that this is a homotopy becomes more clear, even visually. Namely, we see that $\eta$ serves to shift the image of $F$ to the image of $G$ just as a homotopy shifts curves to other curves, and the fact that this shifting jives with the category (i.e. makes the diagram commute) can be interpreted as the necessity of continuity of the shifting in homotopy.

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Let’s take a look at some examples of natural transformations.

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Suppose we fix a group $G$. We claim then that $H\mapsto G\times H$ and $A\xrightarrow{f}B\mapsto A\times G\xrightarrow{f\times1} B\times G$ defines a functor $P_G:\mathbf{Grp}\to\mathbf{Grp}$. Indeed, clearly $P_G(1_A)=1_{A\times G}$ if $A\xrightarrow{j}B\xrightarrow{k}C$ then we see that $P_G(k\circ j)(a,g)=(k(j(a)),g)P_G(k)(j(a),g)=P_G(k)(P_G(j)(a,g))$ and so $P_G(k\circ j)=P_G(k)\circ P_G(j)$. We claim now that any group homorphism $G\xrightarrow{f}H$ induces a natural transformation $\eta:P_G\implies P_H$ with $\eta_A:A\times G\to A\times H$ given by $(a,g)\mapsto (a,f(g))$. Indeed, we must check that for any two groups $A,B$ and any homomorphism $A\xrightarrow{j}B$ the equation $P_H(j)\circ \eta_A=\eta_B\circ P_G(j)$ holds. To do this we just plug and chug: $P_H(j)(\eta_A(a,g)))=P_H(j)((a,f(g))=(j(a),f(g))$ and $\eta_B(P_G(j)(a,g))=\eta_B((j(a),g))=(j(a),f(g))$. Thus, $\eta$ really is a natural transformation. Moreover, it’s clear that if $f$ is an isomorphism then $\eta:P_G\cong P_H$. This then allows us to make such statements as, if $A\cong G, B\cong H$ then $A\times G$ is naturally isomorphic to $B\times H$.

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We now show that “every group is naturally isomorphic to its opposite group , or in our new language the identity functor $1_{\mathbf{Grp}}$ is naturally isomorphic to the opposite functor $F$ which sends $G\mapsto G^{\text{op}}$ and every homomorphism to the associated antihomomorphism defined by $f^{\text{op}}(g)=f(g)^{-1}$. Indeed, for each group $G$ define $\eta_G:G\to G^{\text{op}}$ by $g\mapsto g^{-1}$–this is readily verified to be a group homomorphism. Since $\eta_G$ is, in fact, an isomorphism, to prove it’s a natural equivalence it suffices to show that the naturality condition holds. This amounts to showing that $f^{\text{op}}\circ\eta_G=f$, but this is trivially true.

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As a last classic example, we show that the identity functor is naturally equivalent to the double dual functor on $\mathbf{FinVect}_k$ for some field $k$. Indeed, for each finite dimensional $k$-space $V$ define $\eta_V:V\to V^{\ast\ast}$ by $\eta_V(v)=\text{ev}_v$ where $\text{ev}_v(f)=f(v)$ for $f\in V^\ast$. It is a standard result of linear algebra that since $V$ is finite dimensional, that $\eta_V$ really is an isomorphism for each $V$. Thus, if we can show that $\eta_V$ satisfies the naturality condition, it will automatically be a natural equivalence. To do this we must verify that given any two spaces $V,W$ and any morphism $V\xrightarrow{T}W$ we have that $T^{\ast\ast}\circ\eta_V=\eta_W\circ T$. To do this we pick $v\in V$ and evaluate it at both elements to get $T^{\ast\ast}(\eta_V(v))=T^{\ast\ast}(\text{ev}_v)=\text{ev}_v\circ T^\ast$ and $\eta_W\circ T=\text{ev}_{T(v)}$. To check that these are the same, we pick $f\in W^\ast$ and check that $(\text{ev}_v\circ T^\ast)(f)=\text{ev}_v(f\circ T)=f(T(v))$ and $\text{ev}_{T(v)}(f)=f(T(v))$. Thus, we see that the naturality condition, and thus the fact that the identity functor and double dual functor are naturally equivalent follows.

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We now show that there is a way in which we can compose natural transformations. Indeed, suppose that we have categories $\mathcal{C,D}$ and functors $F,G,H:\mathcal{C}\to\mathcal{D}$. Suppose then that we have natural transformations $\eta:F\implies G$ and $\varepsilon:G\implies H$. Then, we can form the composition $\varepsilon\circ\eta:F\implies H$  by the rule $(\varepsilon\circ\eta)_A=\varepsilon_A\circ\eta_A$ or all objects $A\in\text{obj}(\mathcal{C})$. Note then that a natural transformation $\eta:F\implies G$ is a natural equivalence if and only if there exists a natural transformation $\varepsilon:G\implies F$ such that $\varepsilon\circ\eta=1_F$ and $\eta\circ\varepsilon=1_G$ where, in this context $1_F$ is the map which is identity on each $F(X)$ and likewise for $1_G$.

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References:

[1] Mac, Lane Saunders. Categories for the Working Mathematician. New York: Springer-Verlag, 1994. Print.

[2] Adámek, Jirí, Horst Herrlich, and George E. Strecker. Abstract and Concrete Categories: the Joy of Cats. New York: John Wiley & Sons, 1990. Print.

[3] Berrick, A. J., and M. E. Keating. Categories and Modules with K-theory in View. Cambridge, UK: Cambridge UP, 2000. Print.

[4] Freyd, Peter J. Abelian Categories. New York: Harper & Row, 1964. Print.

[5] Rotman, Joseph J. Introduction to Homological Algebra. Springer-Verlag. Print.

December 28, 2011 -

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