# The Existence of a Natural Isomorphism Confirming the Existence of a Limit, as a Commutativity Diagram

I am reading Category Theory for Programmers and I am having some trouble in Part 2, Chapter 2: Limits and Colimits.

The author writes:

Now that we have two functors, we can talk about natural transformations between them. So without further ado, here’s the conclusion: A functor $D$ from $I$ to $C$ has a limit $\lim D$ if and only if there is a natural isomorphism between the two functors I have just defined: $$C(c, \lim D) ≃ Nat(Δc, D)$$

How can you express the existence of this natural isomophism as a commutative diagram?

• Can you elaborate on what you mean by "as a diagram"? – Hurkyl Sep 22 '17 at 18:40
• @Hurkyl A commutative diagram with objects $I$, $C$ and possibly a topos like $\mathbf{Set}$ for the Hom-objects, and functors as arrows $D$, $\Delta_c$, $C(-,\lim D)$, $Nat(\Delta_-, D)$ such that there's some dotted arrow representing this natural isomorphism which makes the diagram commute if it exists, thus proving the existence of the limit? – Karl Damgaard Asmussen Sep 24 '17 at 8:55

For an object $c\in \mathcal C$, and $D\in\mathcal{C^D}$, what is a natural transformation $\gamma\in\mathrm{Nat}(\Delta c,D)$ like? Like usual, it's going to need to satisfy the commutativity condition $\gamma_{d'}\circ\Delta c(f)=D(f)\circ\gamma_d$ for all $f:d\to d'$ in $\mathcal D$; however, no matter what $f$ is, by definition $\Delta c(f)$ is just $id_c$, so we can simplify that condition to $\gamma_{d'}=D(f)\circ \gamma_d$. But that makes $\gamma$ precisely a cone over $D$. In particular, the limiting cone is just the cone $\alpha:\Delta\circ\lim D\to D$ corresponding to $id_{\lim D}$ by the above isomorphism.
Moreover, given such a $\gamma$, the isomorphism above means it corresponds to a unique morphism $\mathcal C(c,\lim D)$ that is the arrow $f_\gamma$ typically drawn as "the dotted line" on paper. To see why this has the desired property, remember that the isomorphism $\phi:\mathcal C(-,lim D)\to\mathrm{Nat}(\Delta -,D)$ is natural, so that $$\mathrm{Nat}(\Delta(f_\gamma),D)\circ\phi_{\lim D}=\phi_c\circ\mathcal C(f_\gamma,\lim D).$$ If you chase what happens to $id_{\lim D}$ in this square, you'll see that this says exactly that composing the limiting cone with $f_\gamma$ gives you the cone $\gamma$.
• If it makes it easier, just try to use a simple category for $\mathcal D$, like $\mathbb N$ with its usual ordering, or the diagram $\bullet\rightarrow\bullet\leftarrow\bullet$ (which is the shape whose limits are pullbacks). – Malice Vidrine Sep 25 '17 at 17:36