I am trying to prove proposition 11.13 in Adamek’s Joy of Cats. The proposition says that if, in a pullback square, the sink object is a terminal object, then the pullback is a product. I tried to prove it and I wanted to check if it is a right proof.
We assume that the square no. (I) is a pullback square and $T$ here is our terminal object. We assume that $A \times B$ is a product of $A$ and $B$ ($\pi_A$ and $\pi_B$ are corresponding projections – see square no. (II)).
Since $T$ is a terminal object, there must be a unique morphism from $A$, $B$ as well as $A \times B$ to $T$. What we can conclude now? That, $\alpha \circ \pi_A$ is equal to $\beta \circ \pi_B$. This is because $T$ is a terminal object.
We have assumed that $A \times B$ is a product of $A$ and $B$. Hence, there should be a unique morphism from any other source with the same codomain as $A \times B$ to it. $A \times B$ itself (with $A$ and $B$ as its codomain) is one of such sources and hence there should a unique morphism from $A \times B$ to itself. Since $id_{A \times B}$ trivially works here, it must be that unique morphism which makes the whole (II) to commute. In a similar manner, $id_P$ should be the unique morphism that makes the whole (I) to commute.
Now, as we have assumed that $A \times B$ is a product, there should be a unique $m$ making (IV) to commute. In the same way, as we have assumed (I) is a pullback square, there should be a unique $n$ with which (V) commutes.
What we can conclude from all these? see diagram no. (III) above. We can conclude that:
$m \circ n = id_{P}$
$n \circ m = id_{A \times B}$
Therefore $m$ and $A \times B$ are isomorphic. And, this enables us to say P is a product.
A final note, if $T$ was not a terminal object, we could not conclude that (II) is commuting. In this proof, I think, we incorporate this piece of information only once (and that's here).
Thanks
Best Answer
You assume that there is such a thing as a product of $A$ and $B$. On what grounds? All you are given is that $T$ is terminal and a pullback exists. You must then show that the pullback is a product of $A$ and $B$. I do not see in what you have quoted an assumption that you may assume the product exists to begin with... If you assume that $A\times B$ exists, then what you argue seems to be correct. But I don't think that it is warranted to make such an assumption.
In this type of statement, the usual way to proceed is to show that the object in question has the "correct" universal property to be a product for $A$ and $B$. So we have a diagram $$\begin{array}{ccl} &&A\\ &&\downarrow\alpha\\ B&\stackrel{\beta}{\longrightarrow}&T \end{array}$$ where $T$ is terminal. We also have that $$\begin{array}{rcl} P&\stackrel{f}{\longrightarrow}&A\\ g\downarrow&&\downarrow \alpha\\ B&\stackrel{\beta}{\longrightarrow}&T \end{array}$$ is the corresponding pullback diagram. We want to show that $(P,f,g)$ is a product of $A$ and $B$.
So let $X$ be any object, and let $\phi_A\colon X\to A$ and $\phi_B\colon X\to B$ be any two morphisms. We must show that there exists a unique morphism $\Phi\colon X\to P$ such that $\phi_A=\Phi f$ and $\phi_B=\Phi g$.
Because $T$ is terminal, $\alpha\phi_A,\beta\phi_B\colon X\to T$ must be equal. By the universal property of the pullback, there exists a unique $\Phi\colon X\to P$ such that $\phi_A=f\Phi$ and $\phi_B=g\Phi$. But this is exactly what we need.
Thus, $P$ is a product of $A$ and $B$.