Extension of nonisomorphic simple objects

Let $X$ and $Y$ be two nonisomorphic simple objects in an abelian category. Are all extensions of $X$ by $Y$ trivial? ( $\mathrm{Ext}^1(X,Y)=0$ ?)

Pick a field ${\mathbb k}$ and a finite oriented tree $Q$, considered as a category. Then the simple objects in the abelian functor category ${\mathscr A} := k\text{-Vect}^Q$ (which is called the category of representation of $Q$ and is equivalent to the category of modules over the path algebra ${\mathbb k}Q$ of $Q$ over ${\mathbb k}$) are in bijection with the vertices of $Q$, and given two vertices of $Q$, the dimension of $\text{Ext}^1_{\mathscr A}$ between the corresponding simple objects is the number of edges connecting $v$ and $w$.
For example, taking $Q := \bullet\to\bullet$, ${\mathscr A}={\mathbb k}\text{-Vect}^{\bullet\to\bullet}$ is the category of homomorphisms of ${\mathbb k}$ vector spaces (with commutative squares as diagrams), and ${\mathbb k}\xrightarrow{1}{\mathbb k}$ constitutes a (and, up to scalar, the) nontrivial extension between the simples $0\to{\mathbb k}$ and ${\mathbb k}\to 0$.
• finite oriented tree relaced by finite oriented acyclic graph is OK too (if you don't like the word (finite acyclic) quiver). Also, this example is actually the algebra of upper triangular matrix over $\Bbbk$ if one doesn't like quiver representation. The two simple modules correspond to having only entry in $(1,1)$ and $(2,2)$ (with obvious action), then the (unique) non-split extension is associated to the algebra itself. – Aaron Dec 21 '14 at 1:04
• If you have a cycle, then the path algebra becomes infinite dimensional, and representation theory on that algebra may not work with usual quiver representation machinery. For example, $Q$ with one vertex one arrow gives path algebra $\Bbbk[x]$ (the polynomial ring). – Aaron Dec 22 '14 at 20:02