## Franklin's Notes

### Conditions for finite completeness

The purpose of this note is to outline some sufficient/equivalent conditions for a category to be finitely complete , since the proofs of these results are pretty involved.

Proposition 1. In a category $\mathsf{C}$, if $\pi_{xy}$ is a product of $x$ and $y$, and $\pi_{(xy)z}$ is a product of $\pi_{xy}$ and $z$, then $\pi_{(xy)z}$ is a product of $x,y,z$.

Proof. Let $\pi_{xy}$ be the summit of a limit cone $\mu$ with legs $\mu_x,\mu_y$, and $\pi_{(xy)z}$ be the summit of a limit cone $\nu$ with legs $\nu_{\pi_{xy}},\nu_z$. We shall show that $\pi_{(xy)z}$ is a product of $x,y,z$.

Let $\gamma$ be some cone over $x,y,z$ with summit $c$ and components $\gamma_x,\gamma_y,\gamma_z$. Considering only the components $\gamma_x,\gamma_y$ gives a cone over $x,y$, and because $\mu$ is a limit cone of the diagram indexed by $x,y$ with summit $\pi_{xy}$, there exists a unique morphism $f:c\to \pi_{xy}$ which respects the legs of $\mu$, so that $\mu_x f = \gamma_x$ and $\mu_y f = \gamma_y$. Now we have that the morphisms $f:c\to \pi_{xy}$ and $\gamma_z:c\to z$ comprise a cone over $\pi_{xy},z$ with summit $c$, and since $\nu$ is a limit cone over $\pi_{xy},z$, there exists a unique morphism $g:c\to\pi_{(xy)z}$ which respects the legs of $\nu$, so that $\nu_{\pi_{xy}}g=f$ and $\nu_z g=\gamma_z$.

Now, notice that $\pi_{(xy)z}$, together with the three morphisms comprise a cone over $x,y,z$ with summit $\pi_{(xy)z}$. Further, we have found a unique morphism $g:c\to\pi_{(xy)z}$ which respects the legs of $\gamma$ and this cone, since Hence, we have that $\pi_{(xy)z}$ is a limit of the diagram consisting of $x,y,z$, and it is therefore isomorphic to the product of $x,y,z$.

Proposition 2. If a category $\mathsf{C}$ has a terminal object and has all pullbacks, then it has all finite products.

Proof. By the previous proposition, it suffices to show that $\mathsf{C}$ has all binary products, for an n-ary product can be formed as a sequence of $n-1$ binary products.

To see why $\mathsf C$ must have all binary products, let $x,y\in\mathsf C$ be arbitrary objects, and let $t\in\mathsf C$ be a terminal object of $\mathsf C$. Consider the diagram consisting of $f:x\to t$ and $g:y\to t$, where $f,g$ are the unique morphisms from $x,y$ to $t$ respectively. Then, by the existence of pullbacks, there exists $\pi\in\mathsf C$ with morphisms $p_x:\pi\to x$ and $p_y:\pi\to y$ comprising a pullback of this diagram. We shall show that $\pi$ is a product of $x$ and $y$.

Let $c\in\mathsf C$ be an object with some morphisms $j:c\to x$ and $k:c\to y$, so that we have a cone with summit $c$ over the diagram consisting of $x,y$ with no morphisms. If we introduce the unique morphism $l:c\to t$, then we have a cone with summit $c$ over the diagram consisting of $x,y,t$ with the morphisms $f:x\to t$ and $g:y\to t$, for it must be true that $fj=gk=l$ by the uniqueness of morphisms into the terminal object. However, since $\pi$ is a pullback of this diagram, there must exist a unique morphism $p:c\to \pi$ sending the legs $p_x,p_y$ to the legs $j,k$ respectively (any such morphism automatically respects the legs pointing to the terminal object). This shows that $\pi$ is in fact a limit of the diagram consisting only of $x$ and $y$ with no morphisms, and $\pi$ is therefore a product of $x$ and $y$. Since we may form an arbitrary binary product, the proposition is proven. $\blacksquare$