Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
There is no need to assume $\epsilon$ is monic: this is the pullback pasting lemma. – Zhen Lin Aug 21 '13 at 7:21
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.