Better understanding of the pre sheaf kernel of $\varphi$ I'm currently reading chapter II of Hartshorne's algebraic geometry and am comfortable with the definition of a sheaf. I am new to the idea of the presheaf kernel of $\varphi$ however. 
Let $\varphi : \mathscr{F} \to \mathscr{G}$ be a morphism of pre sheaves of abelian groups on a topological space $X$. 
Am I correct in defining the kernel pre sheaf on $\varphi$ to be the contra variant functor that consists of the data 


*

*For every open set $U \subset X$, we have an abelian group $\Gamma(\ker \varphi(U), \mathscr{F})$

*For every inclusion $V \subset U$ open sets in $X$, we have a morphism $\rho_{UV} : \Gamma(\ker \varphi(U), \mathscr{F}) \to \Gamma(\ker \varphi(V), \mathscr{F})$ that satisfies the usual three conditions (empty set, identity map and composition condition)?


Moreover, 
Hartshorne remarks that while the pre sheaf kernel is a sheaf, in general, the pre sheaf cockernel and image are not sheaves. Is there any intuition for why this would be so? 
 A: Your notation is a little bit off. The letter $\Gamma$ typically denotes the global section functor. It takes in a topological space and a sheaf on that space, so $\Gamma(\ker{\varphi}(U), \mathcal{F})$ doesn't make sense. Instead, it's better to define the presheaf kernel $\ker{\varphi}$ as the presheaf that takes an open set $U$ and assigns the abelian group $\ker{\varphi}(U)$, where $\varphi: \mathcal{F} \to \mathcal{G}$ is a morphism of sheaves. The restriction maps, as you observed, should be the restriction maps of $\mathcal{F}$ restricted to $\ker{\varphi}(U)$. If we want to take global section of an open set with respect to this sheaf, we write $\Gamma(U, \ker{\varphi})$.
Now why should this presheaf be a sheaf? Suppose we have an open cover $\{U_i\}$ of our space $X$, and let $s$ be a global section of the presheaf kernel such that $s|_{U_i}=0$ for all $i$. Since the restriction maps of the presheaf are induced by those of the original sheaf, this means that $s|_{U_i}=0$ as a section of $\mathcal{F}$, and so is zero, by virtue of $\mathcal{F}$ being a sheaf.
For the second axiom, suppose we have elements $s_i \in \ker{\varphi}(U_i)$ such that $s_i|_{U_i \cap U_j}=s_j|_{U_i \cap U_j}$ for all $i$ and $j$. Pick $s \in \Gamma(X, \mathcal{F})$ such that $s|_{U_i}=s_i$ for all $i$. To see that $s$ must be in $\ker{\varphi}(X)$, we need to show $\varphi_X(s)=0$. Now since $\varphi$ commutes with the restriction maps, we have $\varphi(s_i)|_{U_i \cap U_j}= \varphi(s_j)|_{U_i \cap U_j}$. Pick $t \in \Gamma(X, \mathcal{G})$ such that $t|_{U_i}=\varphi(s_i)$. Again by commutativity, and uniqueness of sections, we have that $\varphi(s)=t=0$.
The main idea for why the cokernel is not a sheaf is that there's no reason the image of the sheaf should have "enough" sections to satisfy the gluing condition. Notice in the proof above, the essential point is that we're taking sections from our original sheaf.
