1

How can I prove that pullbacks preserve the equivalence of monomorphisms (equivalence in terms of factoring through each other)?

I am reading through a book chapter, and this is one of the left-as-exercise properties that blocks beginners like me from reading further. Specifically, the property is:

If two monomorphisms $f_0:C_0 \to C$, $f_1:C_1 \to C$ are equivalent, i.e. they factor through each other, then their pullbacks $f_0'$ and $f_1'$ in the diagram below are also equivalent monomorphisms.

$$\require{AMScd} \begin{CD} C_0' @> k_0 >> C_0 \\ @V f_0' VV @VV f_0 V \\ C' @>> k > C \\ @A f_1' AA @AA f_1 A \\ C_1' @>> k_1 > C_1 \end{CD}$$

I understand why pullbacks preserve monomorphism. But I can't seem to prove that pullbacks preserve the factors. e.g. for $g$ such that $f_0 = f_1 g$, I can't see an obvious way to map it to a factor $g'$ s.t. $f_0' = f_1' g'$. The construction of such a $g'$ is not obvious to me. I've also tried to build some cones with $id$ in it and use Universal Mapping Property as hinted in the book, but to no avail.

Can someone show how to prove this or give pointers on how to construct such proofs using UMP? (i.e. where to look to construct proofs of such type).

Thanks,

Community
  • 1
tinlyx
  • 1,534

1 Answers1

1

As @Nek pointed out, to prove the equivalence of the pullbacks $f_{0,1}'$:

One can use the fact $kf_1' = f_0(gk_1) $ to construct the following commutative diagram:

$$\require{AMScd} \begin{CD} C_1' @> gk_1 >> C_0 \\ @V f_1' VV @VV f_0 V \\ C' @>> k > C \end{CD}$$

Then by the UMP of the upper pullback,

$\exists! g':C_1' \to C_0'$ such that $g'f_0' = f_1'$,

similarly, one can show

$\exists! h':C_0' \to C_1'$ s.t. $h'f_1' = f_0'$.

Thus the monomorphisms $f_0'$ $f_1'$ factor through each other and are therefore equivalent.

tinlyx
  • 1,534