# Exercise 7.7#

To show that $$i'$$ is an isomorphism, we must show that $$i'⨟q = id_{A'}$$ and $$q⨟i' = id_A$$ for some morphism $$q$$. How do we come up with any morphism $$q: A→A'$$, much less one that is an inverse? If $$A$$ was the vertex of a cone over $$A→B←B'$$ then we could use the universal property of the pullback to generate at least one morphism. It is in fact a cone; the outer square in the following diagram commutes because $$id_A⨟f = f⨟i^{-1}⨟i = f$$:

So we’ve generated some $$u$$ that may be able to serve as the inverse $$q$$ we are looking for. Because of the universal property:

\begin{split} \begin{align} u⨟i' & = id_A \\ u⨟f' & = f⨟i^{-1} \end{align} \end{split}

We can ignore the second equation, but the first equation is half of what we are trying to show. How do we also show that $$i'⨟u = id_{A'}$$? Recall that $$A'$$ is also a cone over $$A→B←B'$$, and in fact the limiting cone:

This means that there is at most one morphism from $$A'$$ to $$A'$$ where the source ($$A'$$) is also a cone over $$A→B←B'$$. The morphism $$i'⨟u$$ exists simply by composition and is also from $$A'$$ to $$A'$$, and so must equal $$id_{A'}$$.

It’s trivial to show the diagram commutes: $$id_A⨟f = f⨟id_B$$. We must also show that for any $$m: C→B$$ and $$n: C→A$$ where the outer square in the following diagram commutes (i.e. $$n⨟f = m⨟id_B$$) there exists a unique morphism $$u: C→A$$ so that $$u⨟f = m$$ and $$u⨟id_A = n$$:

The morphism $$n$$ already satisfies the two equations and so can serve as $$u$$.