Zero, Pullback, Intersection

How do you prove that the intersection of two normal subgroups — or two ideals — is again normal? The naive approach is to verify the condition directly: take $g\in G$$g \in G$, $x\in N\cap M$$x \in N \cap M$, and check that $gx{g}^{-1}\in N\cap M$$gxg^{-1} \in N \cap M$. It works, but it feels like brute force. You are essentially running the same verification twice, once for $N$$N$ and once for $M$$M$, with no insight into why the result is true. The real reason is simpler: normal subgroups and ideals are kernels, and kernels are pullbacks of the zero object...

Let $\mathsf{C}$$\mathsf{C}$ be a category with finite limits. Consider a morphism

$$u:A\to B.$$

We can induce a pair of adjoint functors

$${\mathrm{\Sigma }}_u:\mathsf{C}/A\to \mathsf{C}/B⊣u^{\ast }.$$

Here ${\mathrm{\Sigma }}_u(p:X\to A)=u\circ p:X\to B$$\Sigma_u(p : X \to A) = u \circ p : X \to B$, while $u^{\ast }$$u^*$ is defined via the pullback square. That is, $u^{\ast }(q:Y\to B)=p:X\to A$$u^*(q : Y \to B) = p : X \to A$.

Let us check the adjointness:

$${\mathrm{Hom}}_{\mathsf{C}/B}({\mathrm{\Sigma }}_u(f),g)\cong {\mathrm{Hom}}_{\mathsf{C}/A}(f,u^{\ast }(g)).$$

This is just the universal property of the pullback. Hence $u^{\ast }$$u^*$ preserves limits:

$$u^{\ast }(S{\times }_{B}T)\cong u^{\ast }(S){\times }_A{u}^{\ast }(T).$$

Corollary. Let $\mathsf{C}=\mathsf{Set}$$\mathsf{C} = \mathsf{Set}$, and let $S,T$$S, T$ be two subobjects of $B$$B$. Then

$$u^{\ast }(S{\times }_{B}T)\cong u^{\ast }(S\cap T)\cong u^{\ast }(S){\times }_A{u}^{\ast }(T)\cong u^{\ast }(S)\cap u^{\ast }(T),$$

where $u^{\ast }$$u^*$ is just $u^{-1}$$u^{-1}$.

If $\mathsf{C}$$\mathsf{C}$ has a zero object, then we have $\ker (f,g)=\ker (f)\cap \ker (g)$$\ker(f, g) = \ker(f) \cap \ker(g)$. This follows from the fact that $\ker (f:X\to Y)$$\ker(f : X \to Y)$ is just the pullback of $0\to Y$$0 \to Y$ along $f$$f$.

Now consider $\ker (f,g)=(f,g{)}^{\ast }(0\to Y\times Z)$$\ker(f, g) = (f, g)^*(0 \to Y \times Z)$, where ${\pi }_Y:Y\times Z\to Y$$\pi_Y : Y \times Z \to Y$ and ${\pi }_Z:Y\times Z\to Z$$\pi_Z : Y \times Z \to Z$. We have

$$(0\to Y\times Z)\cong {\pi }_Y^{\ast }(0\to Y){\times }_{Y\times Z}{\pi }_Z^{\ast }(0\to Z).$$

Then

$$(f,g{)}^{\ast }(0\to Y\times Z)\cong (f,g{)}^{\ast }({\pi }_Y^{\ast }(0\to Y){\times }_{Y\times Z}{\pi }_Z^{\ast }(0\to Z))\cong (f,g{)}^{\ast }({\pi }_Y^{\ast }(0\to Y)){\times }_X(f,g{)}^{\ast }({\pi }_Z^{\ast }(0\to Z)),$$

and by pseudofunctoriality of pullback,

$$(f,g{)}^{\ast }\circ {\pi }_Y^{\ast }=({\pi }_Y\circ (f,g){)}^{\ast }=f^{\ast },(f,g{)}^{\ast }\circ {\pi }_Z^{\ast }=({\pi }_Z\circ (f,g){)}^{\ast }=g^{\ast }.$$

Hence

$$(f,g{)}^{\ast }({\pi }_Y^{\ast }(0\to Y)){\times }_X(f,g{)}^{\ast }({\pi }_Z^{\ast }(0\to Z))\cong f^{\ast }(0\to Y){\times }_X{g}^{\ast }(0\to Z),$$

i.e.

$$\ker (f,g)=\ker (f)\cap \ker (g).◼$$