# Abstract Nonsense

## Continuous and Cocontinuous Functors

Point of Post: In this post we discuss the notion of continuous functors as

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Motivation

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Anyone doing math for very long, especially more abstract algebra or topology, is well-acquainted with the fact that constructions “commuting” often makes life very easy. For example, it is a hell of a lot of work to show from first principles that $\pi_1(\mathbb{S}^1)\cong\mathbb{Z}$. After reading such a proof the idea of finding $\pi_1(\mathbb{S}^n)$ for $n>1$ may sound groan-worthy. That is until someone tells you the sigh-inducing fact that $\pi_1(X\times Y)\cong\pi_1(X)\times\pi_1(Y)$. Similarly, trying to compute $\mathbb{Q}^n\otimes_\mathbb{Z}(\mathbb{Z}^r\oplus\mathbb{Z}_{n_1}\oplus\cdots\oplus\mathbb{Z}_{n_m})$ sounds gross until someone points out that $M\otimes(N\oplus L)\cong (M\otimes N)\oplus( M\otimes L)$ and similarly for the other coordinate. In general the ability to commute certain constructions is what enables us to make certain fundamental computations and use these to compute things which are constructed from these fundamental parts.

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Of course, when we say “constructions” we secretly mean some kind of limit or colimit and the things they are commuting with is some kind of functor. Thus, it seems to behoove us, purely for the computational-theorem-proving-goodness involved, to talk about functors which commute with limits (called continuous functors) and those that commute with colimits (called cocontinuous functors).

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Continuous and Cocontinuous Functors

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The basic observation that makes everything else seem like a fairly obvious question is that functors take cones to cones. What exactly does this means? Suppose for a second that we have some diagram $D:\mathcal{I}\to\mathcal{C}$. Then, if we have a functor $F:\mathcal{C}\to\mathcal{D}$ we may compose to get a diagram $F\circ D:\mathcal{I}\to\mathcal{D}$ in $\mathcal{D}$. Moreover, suppose that we have a cone $(C,\{\alpha_i\})$ for $D:\mathcal{I}\to\mathcal{C}$. Just to reiterate, this means that $C$ is an object of $\mathcal{C}$ and $\alpha_i$ arrows $C\to D(i)$ such that $\alpha_j=D(s)\circ\alpha_i$ for every $i\xrightarrow{s}j$. Note then that $F(C)$ is an object of $\mathcal{D}$ and we have arrows $F(C)\xrightarrow{F(\alpha_i)}F(D(i))$ for each $i$ and if $i\xrightarrow{s}j$ then $F(\alpha_j)=F(D(s))\circ F(\alpha_i)$. Thus, we see that $F$ induces a functor (it’s easy to check that it takes compositions to compositions and identity to identity) $\widetilde{F}:\mathbf{Cone}(D)\to\mathbf{Cone}(F\circ D)$ given by $(C,\{\alpha_i\})\mapsto (F(C),\{F(\alpha_i)\})$.

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In particular, we see that $(\lim D,\{\alpha_i\})$ gets mapped to a cone $(F(\lim D),\{F(\alpha_i)\})$–is this a limit of $F\circ D$? Said differently, does the induced functor $\widetilde{F}:\mathbf{Cone}(D)\to\mathbf{Cone}(F\circ D)$ take terminal objects to terminal objects?

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Of course, in general, this is far from true–just take $F$ to be some constant functor. So, when is it true? We call a functor $F$ continuous if this is true for every diagram in $\mathcal{C}$we alternatively say that $F$ commutes with limits.

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Of course, dualizing the above shows that $F$ also induces a functor $\widetilde{F}:\mathbf{Cocone}(D)\to\mathbf{Cocone}(F\circ D)$. It thus makes sense to ask whether $F$ takes colimits to colimits, and we define a functor to be cocontinuous if it does–we alternatively say that $F$ commutes with colimits.

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We define a functor $F$ to be finitely continuous if $F$ commutes with all finite limits, and we define finitely cocontinuous similarly.

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Note that there is a bit of subtlety in the above definitions. Technically, being continuous does not mean that $F(\lim D)=\lim (F\circ D)$ for each diagram $D$. No, while this is often what we really care about it is technically a weaker property than being continuous (see this paper for an example of their non-equivalence).

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While the idea of a functor commuting with limits may sound kind of crazy we actually know tons of examples. Why? Well:

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Theorem: Let $G:\mathcal{C}\to\mathcal{D}$ be a right adjoint functor. Then, $G$ is continuous.

Proof: Let $D:\mathcal{I}\to\mathcal{C}$ be a diagram with limit $(\lim D,\{\alpha_i\})$. We know then that $(G(\lim D),\{G(\alpha_i)\})$ is a cone in $\mathcal{D}$. To see that this cone really is a limit let $(X,\{\beta_i\})$ be another cone for $G\circ D:\mathcal{I}\to\mathcal{D}$. Let $F:\mathcal{D}\to\mathcal{C}$ be the left adjoint of $G$ and $\eta_{X,Y}:\text{Hom}_\mathcal{D}(X,G(Y))\to\text{Hom}_\mathcal{C}(F(X),Y)$ be the guaranteed isomorphism of Hom sets natural in each entry. We know then that since $\beta_i$ is an arrow $X\to G(D(i))$ that $\eta_{X,D(i)}(\beta_i)$ is an arrow $F(X)\to D(i)$. Thus, we see that $\{\eta_{X,D(i)}(\beta_i)\}$ is a collection of arrows $F(X)\to D(i)$–what we claim is that this is a cone. Indeed, we need to check that $s\circ\eta_{X,D(i)}(\beta_i)=\eta_{X,D(j)}(\beta_j)$ for all arrows $i\xrightarrow{s}j$. But, by naturality in the second entry we have that $s^\ast\circ\eta_{x,D(i)}=\eta_{X,D(j)}\circ G(s)^\ast$. Applying the left side of this equality to $\beta_i$ gives

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$(s^\ast\circ\eta{X,D(i)})(\beta_i)=s\circ\eta_{X,D(i)}(\beta_i)$

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and applying to the right gives

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$(\eta_{X,D(j)}\circ G(s)^\ast)(\beta_i)=\eta_{X,D(j)}(G(s)\circ\beta_i)=\eta_{X,D(j)}(\beta_j)$

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where the last equality was reached by using the fact that $(X,\{\beta_i\})$ is a cone for $G\circ D:\mathcal{I}\to\mathcal{D}$. Thus, we see that $(F(X),\{\eta_{X,D(i)}(\beta_i)\})$ really is a cone for $D:\mathcal{I}\to\mathcal{C}$. Now, since $(\lim D,\{\alpha_i\})$ is, well, a limit we get a unique arrow $F(X)\xrightarrow{\nu}\lim D$ such that $\alpha_i\circ\nu=\eta_{X,D(i)}(\beta_i)$ for all $i$. What we note then is that $\eta_{X,\lim D}^{-1}(\nu)$ is an arrow $X\to G(\lim D)$. What we want to be true is that $G(\alpha_i)\circ\eta_{X,\lim D}^{-1}(\nu)=\beta_i$ for all $i$. But, using naturality in the second entry on $\lim D\xrightarrow{\alpha_i}D(i)$ gives

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\begin{aligned}\eta_{X,D(i)}(G(\alpha_i)\circ\eta_{X,D(i)}^{-1}(\nu)) &= (\eta_{X,D(i)}\circ G(\alpha_i)^\ast)(\eta_{X,\lim D}^{-1}(\nu))\\ &= (\alpha_i^\ast\circ\eta_{X,\lim D})(\eta_{X,\lim D}^{-1}(\nu))\\ &= \alpha_i\circ\nu\\ &=\eta_{X,D(i)}(\beta_i)\end{aligned}

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and since $\eta_{X,D(i)}$ is a bijection our desired equality follows. What we must now show is that this is the unique map $X\to G(\lim D)$ for which this equality holds. Suppose that $X\xrightarrow{\tau}G(\lim D)$ is another map such that $G(\alpha_i)\circ\tau=\beta_i$. We see then that $\eta_{X,\lim D}(\tau)$ is an arrow $F(X)\to\lim D$. What we claim is that $\alpha_i\circ\eta_{X,\lim D}(\tau)=\eta_{X,D(i)}(\beta_i)$. But, this is trivial since applying $\eta_{X,D(i)}$ to the left side of $G(\alpha_i)\circ\tau$ clearly (using naturality in the second coordinate again) gives $\alpha_i\circ\eta_{X,\lim D}(\tau)$. By uniqueness of $\nu$ we may conclude that $\eta_{X,\lim D}(\tau)=\nu$ and thus $\tau=\eta_{X,\lim D}^{-1}(\nu)$ as desired. $\blacksquare$

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Of course, dualizing the above gives that

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Theorem: Let $F:\mathcal{C}\to\mathcal{D}$ be a left adjoint functor. Then, $F$ is cocontinuous.

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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] Mitchell, Barry. Theory of Categories. New York: Academic, 1965. Print.

[6] Herrlich, Horst, and George E. Strecker. Category Theory: An Introduction. Lemgo: Heldermann, 2007. Print.