On essential surjectivity of the restriction functor Consider the following setting: $P$ and $Q$ are two finite posets, and $i\colon P\to Q$ is a fully faithful embedding of $P$ in $Q$ (that is, $p\leq p'$ in $P$, if and only if $p\leq p'$ in $Q$). 
Let now $C$ be an additive category (so I have all finite bi-products, but I do not want to assume the existence of other (co)limits) and consider the restriction along $i$ functor:
$$i^*\colon C^Q\to C^P$$
that sends a functor $Q\to C$ to the composition $P\overset{i}{\to} Q\to C$.
Is it possible to prove that $i^*$ is essentially surjective (that is, any diagram $P\to C$ can be extended to a diagram $Q\to C$)? 
Of course if in $C$ there are all finite limits/colimits, then I can construct the right/left Kan extension, that allows me to extend diagrams, but this is not possible without assuming the existence of these limits.
 A: No, you can't do this in general.  To simplify notation, I will consider $P$ to be a subset of $Q$ (so $i$ is the inclusion map).  Now suppose $C$ is a full additive subcategory of an abelian category $D$ and $q\in Q\setminus P$.  Let $F:P\to C$ be a diagram, and let $a$ be the colimit in $D$ of the restriction of $F$ to $\{p\in P:p\leq q\}$ and let $b$ be the limit in $D$ of the restriction of $F$ to $\{p\in P:p\geq q\}$.  There is a natural map $a\to b$, and if we could extend $F$ to $Q$, then this map would factor through $F(q)$.  In particular, if we choose $F$ such that the natural map $a\to b$ is an isomorphism, then $a$ would need to be a direct summand of $F(q)$ in $D$.
So to find a counterexample, we just need to construct a diagram as above such that $a\to b$ is an isomorphism and $a$ is not a direct summand of any object of $C$.  This is not too hard to do.  For instance, let $Q=\{r,s,t,q,x,y,z\}$ where $r\leq s,t\leq q\leq x,y\leq z$, and let $P=Q\setminus\{q\}$.  Then we can just take $a$ to be any object of $D$ which can be expressed as both a pushout of objects of $C$ and a pullback of objects of $C$, but is not a direct summand of any object of $C$.  We define $F:P\to C$ by letting $F|_{\{r,s,t\}}$ be a diagram with pushout $a$ and $F|_{\{x,y,z\}}$ be a diagram with pullback $a$, and then $F$ cannot be extended to $Q$.
For an explicit example of such a $C$, $D$, and $a$, you can take $D$ to be abelian groups and $a=\mathbb{Z}/2\oplus\mathbb{Z}/16$.  Note that $a$ is the pushout of the diagram $\mathbb{Z}/16\leftarrow\mathbb{Z}/4\to\mathbb{Z}/8$ where the maps are multiplication by $4$ and multiplication by $2$, respectively.  Taking Pontryagin duals, $a$ is also the pullback of a diagram $\mathbb{Z}/16\to\mathbb{Z}/4\leftarrow\mathbb{Z}/8$.  So if we take $C$ to be the full subcategory of $D$ consisting of direct sums of $\mathbb{Z}/4$, $\mathbb{Z}/8$, and $\mathbb{Z}/16$, then $a$ is both a pullback and a pushout of objects of $C$, but $a$ is not a direct summand of any object of $C$.
