An example of coherent sheaf I am trying to understand the basics about coherent sheaves (my knowledge of algebraic geometry is rather elementary). Intuitively, to me, they are in a way a weaker notion of than vector bundles. I am trying to look for a simple example that illustrates the difference.
On the wikipedia page here, there is an example of a coherent sheaf on $\mathbb{C}\mathbb{P}^2$ which is not a vector bundle given by
$$\mathcal{O}(1) \xrightarrow{\cdot (x^2-yz,y^3 + xy^2 - xyz)} \mathcal{O}(3)\oplus \mathcal{O}(4) \to \mathcal{E} \to 0$$
i.e. $\mathcal{E}$ is defined as the cokernel. On the zero set defined by $(x^2-yz,y^3 + xy^2 - xyz)$ the first map is just zero so $\mathcal{E}$ should just be $\mathcal{O}(3)\oplus \mathcal{O}(4)$ when restricted to this set, but the wikipedia page says that $\mathcal{E}$ is the zero object. My understanding is that $\mathcal{E}$ should be a line bundle away from this zero set and then the fibre just jumps to a 2 dimensional space  over the zero set but the wikipedia page seems to say that it collapses to a 0 dimensional space. So what is it that I am misunderstanding here?
 A: There are simpler examples that illustrate this subtlety, but I agree and am also confused by what's written on Wikipedia. There are some big differences between how vector bundles and coherent sheaves are used in practice which can be somewhat confusing. For one, vector bundles are usually viewed as geometric objects in themselves whereas coherent sheaves are seen as well-behaved modules in your space. For example, coherent sheaves form an abelian category whereas vector bundles do not, as the following example illustrate.
Let $\mathscr{F}$ be a coherent sheaf on a noetherian scheme $X$ (or a variety $X$ if you're not comfortable with schemes). If $P \in X$ is a point we call the fiber $\mathscr{F}_{(P)}$ of $\mathscr{F}$ at $P$ to be the $k(P) = \mathcal{O}_{X, P}/\mathfrak{m}_P$-vector space $\mathscr{F}_P/\mathfrak{m}_P\mathscr{F}_P$, where $\mathscr{F}_P$ is the stalk of the sheaf $\mathscr{F}$. Note that by the definition of coherence, this is always a finite dimensional space. Remarkably, if $X$ is integral (ie. $\mathcal{O}_{X, P}$ is an integral domain for each $P \in X$), then $\mathscr{F}$ is a vector bundle if and only if the fibers all have the same dimension. (see Hartshorne's Lemma II.8.9)
It is easy to see, though, that if the fibers are not all equidimensional, and $X$ is connected, then $\mathscr{F}$ can not be a vector bundle.
Now, the fiber at $P$ can be viewed as a composition of functors $\mathscr{F} \mapsto \mathscr{F}_P \mapsto \mathscr{F}_P \otimes k(P) = \mathscr{F}_{(P)}$. The first is exact and the second is right exact so that it is a right exact functor of coherent sheaves, to finite-dimensional $k(P)$ vector spaces. Okay, lastly, if $s \in \Gamma(X, \mathscr{F})$ is a global section, we define the vanishing locus of $s$ to be the set $V(s) = \{P \in X\;|\;s_{(P)} = 0 \text{ in } \mathscr{F}_{(P)} \}$.
Now, with all this built up let's do a couple examples. Let $P \in \mathbb{A}^1 $ be the closed point cut out by $(x - \lambda)$ and consider the map $\mathcal{O}_{\mathbb{A}^1} \xrightarrow{\cdot (x - \lambda)} \mathcal{O}_{\mathbb{A}^1}$. Now for all $Q \in \mathbb{A}^1 - \{P\}$, we know that $(x - \lambda)$ is invertible in $\mathcal{O}_{\mathbb{A}^1, Q}$ so this map is an isomorphism there. However, at $P$, the image is exactly $\mathfrak{m}_P$. As such, the cokernel $\mathscr{E}$ of this map, which is a coherent sheaf, is the skyscraper sheaf of $k$ at $P$. In particular, the fiber at $P$ is $k$ and it is zero everywhere else. As such this coherent sheaf is evidently not a vector bundle.
Note: the fiber and the stalk of $\mathscr{E}$ at $P$ agree. Do you see why?
Now let's return to the example from wikipedia and calculate the fibers of $\mathscr{E}$. By the right exactness, the fiber of $\mathscr{E}$ is the cokernel of the map of fibers $\mathcal{O}(1)_{(P)} \to \mathcal{O}(3)_{(P)} \oplus \mathcal{O}(4)_{(P)}$.
In the first case, let $P \in V(x^2 - yz) \cap V(y^3 + xy^3 - xyz)$. Then, $x_2 - yz \in \mathfrak{m}_P\mathcal{O}(2)_P$ and $y^3 + xy^3 - xyz \in \mathfrak{m}_P\mathcal{O}(3)_P$, so that for any $l \in \mathcal{O}(1)_P$, $l(x^2 - yz) \in \mathfrak{m}_P\mathcal{O}(3)_P$ and $l(y^3 + xy^3 - xyz) \in \mathfrak{m}_P\mathcal{O}(4)_P$ so that you're right; the map on fibers is zero when $P$ is in the vanishing locus of the two polynomials.
Similarly, if $P \notin V(x^2 - yz) \cap V(y^3 + xy^3 - xyz)$ then we can show the map $\mathcal{O}(1)_{(P)} \to \mathcal{O}(3)_{(P)} \oplus \mathcal{O}(4)_{(P)}$ is nonzero. This is a bit trickier: we have to fix a line $l \in \Gamma(\mathbb{P}^2, \mathcal{O}(1))$ which does not pass through $P$. Then, since $\mathfrak{m}_P$ is a prime ideal, the image of $l$ under this map is nonzero in at least one of the two summands, so that the map of fibers is a nonzero linear map $k(P) \to k(P) \oplus k(P)$, so that the fiber of the cokernel is always isomorphic to $k(P)$.
Again, you're right that if $U = \mathbb{P}^2 - V$, where $V = V(x^2 - yz) \cap V(y^3 + xy^3 - xyz)$, then $\mathscr{E}|_U$ is indeed a line bundle since all the fibers are one dimensional.
