Johnstone's Topos Theory, Lemma 0.17 - the lower-right square is a pushout, proof help.

Question

Here is work I did so far on the proof, with more in depth explanation than what the book has.

1. Every equalizer arrow in a category is a monomorphism, so that dually, every coequilizer arrow is an epimorphism.

2. $\gamma_3 = \operatorname{coeq}(\gamma_1, \gamma_2)$ by assumption (see "rows and columns are coequilizers").

3. Since $f_3$ is a coequilizer, it is an epimorphism. We need to prove that $\gamma_3$ is still a coequilizer but of $\gamma_1 f_3, \gamma_2 f_3$. But $f_3$ is an epimorphism so if $\gamma_3 \gamma_1 f_3 = \gamma_3 \gamma_2 f_3$, then we automatically get $\gamma_3 \gamma_1 = \gamma_3 \gamma_2$, by definition of epimorphism. And conversely, since $\gamma_3 \gamma_1 = \gamma_3 \gamma_2$ by definition of coequilizer $\gamma_3$, we automatically get $\gamma_3 \gamma_1 f_3 = \gamma_3 \gamma_2 f_3$. We have left to prove that if $\gamma_3' \gamma_1 f_3 = \gamma_3' \gamma_2 f_3$ for some other morphism $\gamma_3'$, then there is a unique morphism $u$ such that $\gamma_3' = u\gamma_3$. We already have $u$ though from $\gamma_3 = \operatorname{coeq}(\gamma_1, \gamma_2)$ and so $u$ is by assumption unique. In summary we have $\gamma_3 = \operatorname{coeq}(\gamma_1 f_3, \gamma_2 f_3)$ as well!

4. By commutativity of the diagram, in particular the upper-right square commutes, which means $\gamma_1 f_3 = g_3 \beta_1$ and $\gamma_2 f_3 = g_3 \beta_2$, i.e. they commute along common indices. (Note there is a typo in the book, so just take commutativity w.r.t. parallel arrows to mean commutative along common indices). Thus since the two compositions (in each index $i=1,2$) are equal, we can simply substitute into the "coequilizer equation" we arrived at in the previous step to get: $\gamma_3 = \text{coeq}(g_3 \beta_1, g_3\beta_2)$.

5. Next, we want to prove that the lower-right square is a pushout. Let $Z_3'$ be another object and $Z_2 \xrightarrow{h_3'} Z_3', \ Y_3 \xrightarrow{\gamma_3'} Z_3'$ be two morphisms such that $h_3' \beta_3 = \gamma_3' g_3$. Then by $\beta_3 = \operatorname{coeq}(\beta_1, \beta_2)$ by assumption, we have the following chain of equalities: $$ \gamma_3' g_3 \beta_1 =h_3'\beta_3\beta_1 = h_3'\beta_3\beta_2 = \gamma_3'g_3 \beta_2 \tag{5} $$ But by the result of step 4, we have that $\gamma_3 = \operatorname{coeq}(g_3 \beta_1, g_3\beta_2)$ so that there must exist a unique morphism $u$ such that $\gamma_3' = u \gamma_3$. For the square to be truely a pushout though we also need that $h_3' = uh_3$.

The part I'm having trouble with is showing that $h_3' = u h_3$ as well. That would conclude that the lower-right square is indeed a pushout, but there are still a few more steps left in the proof. I will make another question for those if I should have trouble with them.

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I don't know if we'll use it, but I just came up with:

The problem's commutative diagram pictured is symmetric across its diagonal (together with corresponding properties), so we must also have: $h_3 = \operatorname{coeq}(\beta_3 g_1, \beta_3 g_2)$ by the exact same argument used in step 3.

Answer 1

Let $u$ be the unique pushout morphism such that $\gamma_3' = u \gamma_3$, so that by definition of $h_3', \gamma_3'$ we have that $\gamma_3' g_3 = u \gamma_3 g_3 = h_3' \beta_3$. Let $v$ be the unique map such that $h_3' = v h_3$ since $h_3 = \operatorname{coeq}(\beta_3 g_1, \beta_3 g_2)$.

But $v\gamma_3 g_3 = v h_3 \beta_3 = h_3' \beta_3 = \gamma_3' g_3 = u \gamma_3 g_3$. All equalities are by assumed commutativity of the various diagrams.

So that by $g_3$ being an epimorphism we have: $v\gamma_3 = u \gamma_3$, and by $\gamma_3$ being an epimorphism we have $u = v$.

That completes the proof that the lower-right square in Lemma 0.17 of Johnstone's book is indeed a pushout square.