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Reading Hungerford's Algebra I encountered the following statement

Consider the diagram $$\require{AMScd} \begin{CD} A @>{\alpha}>>B @>{\beta}>> C\\ @V{\gamma}VV @V{\delta}VV @V{\epsilon}VV\\ D @>{\zeta}>> E @>{\eta}>> F \end{CD}$$ where $(A,\alpha,\gamma)$ is a pullback of $\zeta,\delta$ and likewise $(B,\beta,\delta)$ is a pullback of $\eta,\epsilon$. If $\epsilon$ is monic, then the outer rectangle is also a pullback.

It would seem to me that the assumption that $\epsilon$ be monic is ancillary. That the outer square commutes is trivial; suppose we have $(Z,\theta,\kappa)$ such that $$\require{AMScd} \begin{CD} Z @>{\kappa}>>C\\ @V{\theta}VV @V{\epsilon}VV \\ D @>{\eta \circ \zeta}>> F \end{CD}$$ is commutative. Then by universal property of $(B,\delta,\epsilon)$ we have a unique morphism $t: Z \to B$ such that $\delta \circ t = \zeta \circ \theta$ and $\beta \circ t= \kappa$.

Repeating the above except comparing $(A,\alpha,\gamma)$ to $(Z,\theta,t)$, we obtain a unique morphism $s: Z \to A$ such that $\alpha \circ s = t$ and $\gamma \circ s = \theta$. Then $$ \beta \circ t = \beta \circ \alpha \circ s = \kappa $$ and as $\gamma \circ s = \theta$ it follows that $s$ is indeed the desired morphism.

Can anyone comment on the integrity of the proof, and/or the necessity of the hypothesis?

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There is no need to assume $\epsilon$ is monic: this is the pullback pasting lemma. –  Zhen Lin Aug 21 '13 at 7:21
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A diagrammatic proof of the pullback lemma is on ProofWiki. –  Lord_Farin Aug 21 '13 at 13:27
    
Perhaps I am misunderstanding the ProofWiki entry, but what in their argument avoids the issue pointed out in the Santiago Canez's answer? –  Cameron Aug 22 '13 at 3:17

1 Answer 1

After you obtain $s$, you only know it is the unique morphism satisfying $\alpha \circ s = t$ and $\gamma \circ s = \theta$, but in order for the outer rectangle to be a pullback you need to know that it is the unique morphism satisfying $\beta \circ \alpha \circ s = \kappa$ and $\gamma \circ s = \theta$. Note that $\alpha \circ t$ implies $\beta \circ \alpha \circ s = \kappa$ as you explain, but not conversely so you cannot go from $\beta \circ \alpha \circ s' = \kappa$ to $\alpha \circ s' = t$ in order to use the uniqueness you do know to conclude that $s=s'$.

Here's a hint: show that if $\epsilon$ is monic, so is $\delta$.

Edit: As pointed out in a comment, you don't actually need $\epsilon$ to be monic.

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It can be shown that if $\epsilon$ is monic, it follows that $\gamma$ is monic. Hence for any $s'$ satisfying $\gamma \circ s' = \theta$ we have $\gamma \circ s' = \gamma \circ s = \theta$ and therefore $s = s'$. But from what I see on nlab and proofwiki there seems to be another way. –  Cameron Aug 21 '13 at 20:24

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