Isomorphism between (source of) kernels of parallel arrow of a pullback square, by adjunction

Question

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"?

Answer 1

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 $X \to S$.