If $G : \mathcal{D} \to \mathcal{C}$ has a left adjoint then $G$ preserves kernels.

Question

The theorem 7.7 in Course an homologic algebra of Peter Hilton is

If $G: \mathcal{D} \to \mathcal{C}$ has a left adjoint then G preserves products, pull-backs and kernels.

Hilton prove that $G$ prerserves products and pull backs and leave the proof that $G$ preserv kernel as exercise. How can I prove that $ G $ preserves the kernel? There are some demonstrations with notions of limits and colimites, but they are notions that are not used in the book. I want to try it without using the concepts of limits and colimites. How can I do it?

Answer 1

$\mathcal{C}$ and $\mathcal{D}$ should admit zero object. Let $\eta: F \dashv G$.

We are going to show two things:

1. $G$ preserves zero objects hence zero morphisms.
2. $G$ preserves equalizers, i.e. if $f$ is the equalizer of $\varphi_1$ and $\varphi_2$ then $Gf$ is the equalizer of $G\varphi_1$ and $G\varphi_2$.

It's easy to see that $f$ is the kernel of $\varphi$ iff $f$ is the equalizer of $0$ and $\varphi$. Then the original statement is proved.

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Proof of 1.

Let $Z$ be the zero object in $\mathcal{D}$ and $N$ an object in $\mathcal{C}$. Since $\eta : Hom(FN,Z) \rightarrow Hom(N,GZ)$ is bijective, $\left| Hom(N,GZ)\right| = \left| Hom(FN,Z)\right| = 1$. Therefore $GZ$ is terminal object hence zero object. It immediately follows that $G$ preserves zero morphisms.

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Proof of 2.

Let $f$ be the equalizer of $\varphi_1$ and $\varphi_2$. For any $\tau$ such that $G\varphi_1 \circ \tau = G\varphi_2 \circ \tau$, we have $\varphi_1 \circ \eta^{-1}\tau = \varphi_2 \circ \eta^{-1}\tau$ by naturality of $\eta^{-1}$. So there is a unique morhism $\tau_0$ such that $f \circ \tau_0 = \eta^{-1}\tau$. Again, by bijectivity and naturality of $\eta$, $\eta\tau_0$ is the unique morphism such that $Gf \circ \eta\tau_0 = \tau$, so $Gf$ is the equalizer of $G\varphi_1$ and $G\varphi_2$.