Functor preserves kernels iff it's left exact

Question

I'm trying to understand the proof to a statement in Rotman's 'Introduction to Homological Algebra': Proposition 5.25, p. 240:

Let $F :_R\text{Mod} \to \text{Ab}$ be a covariant functor. Then $F$ preserves kernels iff $F$ is left exact.

It's my understanding that in this context, kernel means a solution to a pullback (where one of the objects is zero), or inverse limit if we construct the pullback definition in terms of the appropriate inverse system.

Here's the proof that's given:

Note: Example 5.12 (iii) just shows how the kernel can be seen as a pullback of a particular system.

Let $0 \to A' \xrightarrow i A \xrightarrow p A''$ be exact. If $F$ preserves kernels, then $(FA', Fi)$ is a kernel of $Fp$, and Example 5.12 (iii) shows that $Fi$ is an injection, that is, $F$ is left exact. Conversely, if $F$ is left exact, then $(\text{ker } Fp, Fi)$ is a kernel of $Fp$ so $F$ preserves kernels.

I don't see how either direction of his proof shows what was stated. If someone could fill in the details of what I'm not seeing, I'd appreciate it.

I was able to show the first implication, but with considerably more work than what was stated as a proof.

For the first implication, we assume $0 \to A' \xrightarrow i A \xrightarrow p A''$ is exact. I want to show that $0 \to FA' \xrightarrow{Fi} FA \xrightarrow{Fp} FA''$ is exact which means I need to show three things: $\text{ker }(Fi) = 0$, $\text{im }(Fi) \subseteq \text{ker }(Fp)$ and $\text{ker }(Fp) \subseteq \text{im }(Fi)$.

1. Clearly, we can show that $(A', i)$ is a kernel of $p : A \to A''$, and so $(FA', Fi)$ is a kernel of $Fp : FA \to FA''$ by hypothesis. This tells us one of the three things we needed: $Fp \circ Fi = 0$ which means $\text{im }(Fi) \subseteq \text{ker }(Fp)$.
2. He say's "Example 5.2 (iii) shows that $Fi$ is an injection", but that's not really immediate. All Example 5.2 (iii) says it that the kernel of a function is a pullback. In this case $(0, z)$ is a kernel of $i : A' \to A$ where $z : 0 \to A'$ is just the zero map. Since, by hypothesis, $F$ preservies kernels, we have that $(F0, Fz)$ is a kernel of $Fi : FA' \to FA$. In particular, we have $\text{ker } Fi \cong F(0) = 0$ because kernels are unique up to isomorphism, which tells us $Fi$ is injective.
3. As $(FA', Fi)$ is a kernel of $Fp : FA \to FA''$, we can use the definition of pullback or equalizer and choose $X := \text{ker } (Fp)$ and $h : \text{ker }(Fp) \to FA$ to be the inclusion. Clearly $Fp \circ h = 0$ so there exists $\theta : \text{ker }(Fp) \to FA'$ such that $Fi \circ \theta = h$, in particular for $x \in \text{ker }(Fp)$, we have $x = h(x) = Fi \circ \theta(x)$ which means $x \in \text{im }(Fi)$.

The converse also requires quite a bit more work than is stated. Am I missing something , or is it expected to have to go through all these details to arrive at the same conclusions? My concern is that I'm missing something that I shouldn't be that makes my life much easier? I'll add my work in any case.

Suppose $F$ is left exact and let $p : A \to A''$ be a morphism and $(K, i)$ be it's kernel. It can be shown that $i$ is injective because $\text{ker }(p)$ is also it's kernel and kernels are unique up to isomorphism. We can just say: without loss of generality, $K = \text{ker }(p)$ and $i$ is the inclusion map. So we have an exact sequence $0 \to K \xrightarrow i A \xrightarrow p A''$. By hypothesis we have $0 \to FK \xrightarrow{Fi} FA \xrightarrow{Fp} FA''$ is exact. We want to show that $(FK, Fi)$ is a kernel for $Fp : FA \to FA''$. We know that $(\text{ker }(Fp), j)$ where $j : \text{ker }(Fp) \to FA$ is the inclusion map is a kernel of $Fp : FA \to FA''$. Using the definition of pullback we can choose $X := Fk$ and $h = Fi$ with $Fp \circ h = 0$. This means that there exists $\theta : FK \to \text{ker }(Fp)$ such that $j \circ \theta = Fi$. We can show that $\theta$ is an isomorphism, which means that $FK$ is a kernel of $Fp : FA \to FA''$ as desired.

Answer 1

You actually have to require that $F$ is an additive functor, then Rotman's proof says it all.

About the first implication, you should maybe first clearly understand that

i) if $A' \xrightarrow i A$ is a mono, then $A' \cong Im(i)$ (try to prove it directly);

ii) if $0 \to A' \xrightarrow i A \xrightarrow p A''$ is exact, then $i = ker(p)$ (also easy to prove just using the definitions);

Then you have that $ KerF(i) \cong F(Ker(i)) \cong F(0) \cong 0 $ (using here functor's additivity). So $F(i)$ is mono and $KerF(p) \cong F(Ker(p)) \cong F(Im(i)) \cong F(A') \cong ImF(i)$.

This means that $0 \to F(A') \rightarrow F(A) \rightarrow F(A'')$ is exact.

Answer 2

One useful relevant fact:

$$0 \to A \xrightarrow{f} B \xrightarrow{g} C$$ is exact iff $A \xrightarrow{f} B$ is a kernel of $B \xrightarrow{g} C$.