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Stream: learning: questions

Topic: Left direction of the pullback lemma

Olli (Dec 28 2020 at 10:39):

I'm having trouble proving the left direction of the pullback lemma. First I'll write how I proved the right direction with morphism as below:
image.png

To prove the $\Rightarrow$ part (R and L being PB implies that L+R is PB):

• Suppose $fmi = gj$. To show that L+R is a pullback, we need to show that there exists a unique $z$ such that $fmi = fmnz$ and $gj = gkez$.
• Because $fmi = gj$, by the UP of the right pullback, $\exists!y : fmi = fhy = gky = gj$.
• Because $mi = hy$, by the UP of the left pullback $\exists!x : mi = mnx = hex = hy$.

• Take $x$ as $z$:

  1. Because $mi = mnx$ then $fmi = fmnx$.
  2. Because $fh = gk$ (R commutes by assumption) and $hy = hex$, we can combine this as $gj = fhy = fhex = gkex$ and we are done.

Now continuing with the following:
image.png

To prove the $\Leftarrow$ direction (L+R and R being PB implies that L is PB):

• Suppose $mi = hj$. Our goal is to show $\exists!x$ such that $mi = mnx$ and $hj = hex$.
• Because $fmi = gkj$, by UP of R being a PB, $\exists!y : fmi = fhy = gky = gkj$.
• Because $fmi = gkj$, by UP of L+R being a PB, $\exists!z : fmi = fmnz = gkez = gkj$.

• Take $z$ as $x$:

  1. From $fmi = fmnz$ I need to conclude that $mi = mnz$, which is true if $f$ is monic, but I don't have that as an assumption. In the $\Rightarrow$ direction there is no issue because we are just applying $f$ to things that are equal.
  2. If I just read from the diagram I can see that $hj = hy = hez$, and I have that $he = mn$ by assumption, so again if I could conclude from $fmi = fmnz$ that $mi = mnz$, then I could get $hj = mi = mnz = hez$, but this is of course the same problem as in 1. I think for a proper proof I need something more than "just follow the diagram", but I think I am missing an obvious assumption in my chasing attempt.

Olli (Dec 28 2020 at 10:49):

Whops, the $b$ in the picture above is supposed to be $j$

Ralph Sarkis (Dec 28 2020 at 11:14):

I think you do not have the right definition of pullback. For instance, in the first direction, you say

Suppose $fmi = gj$. To show that L+R is a pullback, we need to show that there exists a unique $z$ such that $fmi = fmnz$ and $gj = gkez$.

But that is not enough, you should show $z$ is unique such that $nz = i$ and $kez = j$.