Is the kernel of projections in an additive category preserved by additive functors? Let $\mathcal{C}$ and $\mathcal{D}$ be an additive categories and let $\psi : \mathcal{C} \to \mathcal{D}$ be an additive functor. I have two questions : 

  
*
  
*Is it true that if $f: C \to C$ is a projection in $\mathcal{C}$ which has a kernel, then $1-f$ too has a kernel ?
  
  
  
*Is it true that : With the above setup, if a projection $f$ has a kernel, then so does $\psi(f)$, and $\psi(ker f) \simeq ker (\psi(f))$ ?
  
  

There is a canonical map  $\psi(ker f) \to ker (\psi(f))$, but I do not know how to show it is an isomorphism. 
I understand that additive functors which preserve kernels are called left-exact and that not all functors are left exact, but I have a feeling (possibly incorrect) that additive functors preserve kernels of projections. I would appreciate any hints. Thanks 
 A: Regarding question 2, if $f$ has a kernel $i : K \to C$, then the fact $f \cdot (1-f) = 0$ implies there is a map $p : C \to K$ with $ip = 1-f$ and $pi = 1$.
Note that a kernel of $f$ is the same thing as an equalizer of $1-f$ and $1$. 
This sort of arrangement is important not just in additive categories, but any category at all. For example,
Theorem: If $p : C \to K$ and $i : K \to C$ are arrows satisfying $pi = 1$, then $i$ is an equalizer of $ip$ and $1_X$. Moreover, given any $f : D \to C$ with $ip \cdot f = 1_C \cdot f$, the corresponding map $ D \to K$ given by the equalizer property is $pf$.
The property $pi=1$ is preserved by any functor at all, and this is an absolute equalizer: it is preserved by all functors.

Regarding your first question, I'm pretty sure the answer is no. Take the preadditive category generated by


*

*Two objects $K$ and $C$

*An arrow $p : C \to K$

*An arrow $i : K \to C$

*A relation $pi = 1$.


In particular, the homsets are:


*

*$\hom(K, K) \cong \mathbf{Z}$, generated by $1$

*$\hom(K, C) \cong \mathbf{Z}$, generated by the i

*$\hom(C, K) \cong \mathbf{Z}$, generated by the p

*$\hom(C, C) \cong \mathbf{Z}^2$, generated by $ip$ and $1$


If you extend this to an additive category by adding in all direct sums, I'm pretty sure you will never include the complementary summand to $K$. I don't have a snazzy way to prove it off hand.
