Let $\mathcal A$ be an abelian category, $\alpha_1 \colon A_1 \to B$, $\alpha_2 \colon A_2 \to B$ two morphisms and $A_1 \leftarrow A_1\times_B A_2 \to A_2$ their pullback (with morphisms, say, $p_1, p_2$).

I can easily show, by hands, that the unique morphism $\varphi\colon\text{Ker}\, p_2 \to \text{Ker}\,\alpha_1$ induced by the universal property of $\ker\alpha_1 \colon\text{Ker}\,\alpha_1 \to A_1$ is an isomorphism.

I have been told that, actually, this is because "limits preserve limits" (or "limits commute with limits"). I know that the limit functor $\lim\colon\mathcal{A^D}\to A$ is right-adjoint to the costant functor $\Delta\colon \mathcal{A}\to \mathcal{A^D}$, but I cannot see how to prove the above result using it.

Can you provide any reference, also in order to become familiar with this "limits twists"?


The phrase "limits commute with limits" means the following: Let $C$ be a complete category and $I,J$ small categories. If $F : I \times J \to C$ is a functor, then there are canonical isomorphisms $\lim_i \lim_j F(i,j) \cong \lim_{i,j} F(i,j) \cong \lim_j \lim_i F(i,j)$. A proof can be found in the book by Mac Lane, but basically it is an easy exercise playing around with universal properties. Nothing really happens.

As a special case of this, one gets the canonical isomorphism $(X \times_S S') \times_{S'} T \cong X \times_S T$ in an arbitrary category where these fiber products make sense (see also MO/80797). If $T$ is a zero object, this means that the kernel of $X \times_S S' \to S'$ is the same as the kernel of $X \to S$.

  • $\begingroup$ Very clever! Way easier than I was thinking about. Actually, I was also trying to prove that "composition of pullbacks is a pullback" by mean of commutativity of limits. So, double answer! Thank you, Martin (once more!) $\endgroup$ – Andrea Gagna Feb 20 '13 at 1:52
  • $\begingroup$ Actually, isn't $(X \times_S S') \times_{S'} T \cong X \times_S T$ the pullback pasting lemma? This isomorphism has a somewhat different flavour than, say, $\ker (f \times g) \cong \ker (f) \times \ker (g)$. $\endgroup$ – Zhen Lin Feb 20 '13 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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