# Exercise 7.4#

Taking morphism labels from Wikipedia:

We’ll take rt to mean r∘t, as in the Wikipedia article, for brevity. We’ll break the problem into part 1. where we show that if the left square is a pullback then the whole rectangle is a pullback and part 2. where we show the converse.

For part 1., if the left square is a pullback, then for any other cone $$(Q,q_1,q_2)$$ over $$A'→B'←B$$ there is some unique morphism $$k: Q→A$$ such that $$k$$ factors out of the pair of maps $$q_2: Q→B$$ and $$q_1: Q→A'$$. That is, we have that $$q_2 = t∘k$$ and $$q_1 = u∘k$$. The $$Q,q_1,q2$$ variable names are coming from Pullback (category theory).

Take some arbitrary (in the sense of ∀, see Glossary of mathematical jargon § Descriptive informalities) third morphism $$q_3: Q→C$$ where $$f∘q_3 = g∘h∘q_1$$ to come up with an arbitrary cone $$(Q,q_1,q_3)$$ over $$A'→C'←C$$. A picture to have in mind is:

Is there always a unique morphism from $$(Q,q_1,q_3)$$ to $$(A,rt,u)$$? We know that $$k$$ is a good candidate because $$q_1 = u∘k$$. Is it also always the case that $$q_3 = l∘k$$ for some morphism $$l$$?

Because the right square is a pullback, $$(B,r,s)$$ is the limit with respect to the cospan $$B'→C'←C$$. We know that $$(Q,h∘q_1,q_3)$$ is also a cone over $$B'→C'←C$$, so we have that $$q_3 = r∘q_2$$. Given $$q_2 = t∘k$$ (see above) we then have that $$q_3 = r∘t∘k$$ and so $$l = r∘t$$ and $$k$$ can serve as the unique morphism from $$(Q,q_1,q_3)$$ to $$(A,rt,u)$$. This makes the limit of f and gh equal $$(A,rt,u)$$, which is what we wanted to show.

For part 2. we assume that the whole rectangle is a pullback and must show that the left rectangle is a pullback as well.

Because the whole rectangle is a pullback, then for any other cone $$(Q,q_1,q_3)$$ over $$A'→C'←C$$ there is some unique morphism $$k: Q→A$$ such that $$k$$ factors out of the pair of maps $$q_3: Q→C$$ and $$q_1: Q→A'$$. That is, we have that $$q_3 = r∘t∘k$$ and $$q_1 = u∘k$$.

Take some arbitrary third morphism $$q_2: Q→B$$ where $$s∘q_2 = h∘q_1$$ to come up with an arbitrary cone $$(Q,q_1,q_2)$$ over $$A'→B'←B$$. Is there always a unique morphism from $$(Q,q_1,q_2)$$ to $$(A,t,u)$$? We know that $$k$$ is a good candidate because $$q_1 = u∘k$$. Is it also always the case that $$q_2 = m∘k$$ for some morphism $$m$$?

Because the right square is a pullback, $$(B,r,s)$$ is the limit with respect to the cospan $$B'→C'←C$$. We know that $$g∘h∘q_1 = f∘q_3$$ and we have that $$g∘h∘q_1 = g∘s∘q_2$$ because $$h∘q_1 = s∘q_2$$, so $$g∘s∘q_2 = f∘q_3$$. So $$(Q,s∘q_2,q_3)$$ is also a cone over $$B'→C'←C$$, so we have that $$q_3 = r∘q_2$$. Given $$q_3 = r∘t∘k$$ (see above) we then have that $$q_2 = t∘k$$ so $$m = t$$ and $$k$$ can serve as the unique morphism from $$(Q,q_1,q_2)$$ to $$(A,rt,u)$$. This makes the limit of s and h equal $$(A,t,u)$$, which is what we wanted to show.