I'm trying to deal with an example of a rank two vector bundle over the complex projective plane which is non slope-stable (because the associated sheaf of sections has a coherent subsheaf of equal slope) but it does not admit subbundles with greater slope. This is the simplest example I have in mind in order to explain why in dimension greater than 1 (of the Kahler manifold) one has to deal with subsheaves and not only subbundles when checking the stability of a vector bundle. I have problems in many part of the exposition which goes like this:


Let $p\in\mathbb{CP}^2$ be a point and $\mathcal{I}\_p$ its ideal sheaf. Then, by using the Koszul resolution, we have the following short exact sequence of coherent sheaves $0\rightarrow\mathcal{O}(-2)\overset{f}{\rightarrow}\mathcal{O}(-1)\oplus\mathcal{O}(-1)\rightarrow\mathcal{I}\_p\rightarrow 0$. Now take any non trivial sheaf homomorphism $\mathcal{O}(-2)\overset{g}{\rightarrow}\mathcal{O}$, which exists since it can be seen as a holomorphic section of $\mathcal{O}(-2)^\*\otimes\mathcal{O}\cong\mathcal{O}(2)$, and let $\mathcal{E}$ be the push-out of $f$ and $g$ in the catgeory of coherent sheaf.


**Question 1:** How can we prove that $\mathcal{E}$ fits in the following sequence of sheaves $0\rightarrow\mathcal{O}\rightarrow\mathcal{E}\rightarrow\mathcal{I}\_p\rightarrow 0$ ? Why is it exact?


*I have only understood that the first arrow is injective since $f$ it is.*


By using this exact sequence we know that $\mu(\mathcal{E}):=\frac{deg(\mathcal{E})}{rk(\mathcal{E})}=\frac{0}{2}=0$ but since $\mu(\mathcal{O})=0$ we conclude that $\mathcal{E}$ is not stable.


**Question 2:** Why $\mathcal{E}$ does not admit subbundles of greater slope?


**EDIT** *As Libli has wonderfully explained, this example shows that stability should be checked also on quotient sheaves and not only on quotient bundles. Moreover one has to prove that such a push-out $\mathcal{E}$ is indeed locally-free and not only coherent (as Libli has done).*


**EDIT 2** *If you are interested in this example you can find it in the wonderful book of Huybrechts-Lehn "The Geometry of Moduli Spaces of Sheaves" (Thm 5.1.1 and Ex. 5.1.2)*



Push-out in the category of coherent sheaves over the complex projective plane
This example is useful not to understand that stability should be checked on subsheaves (instead of checking only subbundles), but that it should be checked for **all quotient sheaves** (and not only quotient bundles). Indeed, let us show that $\mathcal{E}$ has no quotient bundle of rank $1$ with slope $0$. Assume, by absurd, it has. This means, we have an exact sequence:


$$0 \longrightarrow \mathcal{F} \longrightarrow \mathcal{E} \longrightarrow \mathcal{O} \longrightarrow 0.$$


By the local duality for locally complete intersections, we have $\mathcal{H}om(\mathcal{I}\_p, \mathcal{O}) = \mathcal{O}$, $\mathcal{E}xt^1(\mathcal{I}\_p, \mathcal{O}) = \mathcal{O}\_p$ and $\mathcal{E}xt^i(\mathcal{I}\_p, \mathcal{O}) = 0$ for all $i>1$. We deduce that $\mathcal{E}xt^i(\mathcal{E}, \mathcal{O}) = 0$ for all $i>0$. As a consequence, we have $\mathcal{E}$ is a rank $2$ vector bundle on $\mathbb{P}^2$ (you didn't ask that question, but it is certainly not obvious to see that $\mathcal{E}$ is actually a vector bundle with the *painful* push-out construction).


In particular, $\mathcal{F}$ would be a reflexive sheaf of rank $1$, that is a line bundle on $\mathbb{P}^2$. Since its slope is $0$, we have $\mathcal{F} = \mathcal{O}$. But $$\mathrm{Ext}^1(\mathcal{O},\mathcal{O}) = H^1(\mathbb{P}^2,\mathcal{O}) = 0,$$ so that $\mathcal{E}$ would be the direct summ $\mathcal{O} \oplus \mathcal{O}$. This is certainly not true, as they are no exact sequence:


$$ 0 \longrightarrow \mathcal{O} \longrightarrow \mathcal{O} \oplus \mathcal{O} \longrightarrow \mathcal{I}\_p \longrightarrow 0$$ on $\mathbb{P}^2$. Hence, if we only look at quotient bundles of $\mathcal{E}$, one could have the wrong impression that $\mathcal{E}$ is slope-stable. This is not true as $\mathcal{E}$ has a destabilizing quotient sheaf of rank $1$, namely $\mathcal{I}\_p$.


