TITLE: Simple objects with zero dimensional support
QUESTION [5 upvotes]: I am reading the book “Fourier-Mukai transforms in algebraic geometry” by Daniel Huybrechts. Before stating my question, let me state some definitions. Here, $X$ is a smooth projective variety.
Definition: In a k-linear triangulated category $\mathcal{D}$, an object $P\in \mathcal{D}$ is called simple if $Hom(P,P)$ is a field.
Definition: The support of a complex $\mathcal{F}^{\bullet}\in D^b(X)$ is the union of all its cohomology sheaves, i.e. it is the closed subset
$$supp(\mathcal{F}^{\bullet}):=\cup \ supp(H^i(\mathcal{F}^{\bullet}))$$.
And for support of complexes, we have the following lemma (it is stated on page 65 as Lemma 3.9 in Huybrecht’s book):
Lemma: Suppose $\mathcal{F}^{\bullet}\in D^b(X)$ and $supp(\mathcal{F}^{\bullet})=Z_1\cup Z_2$ where $Z_1,Z_2$ are disjoint closed subsets of $X$. Then, $\mathcal{F}^{\bullet}\cong \mathcal{F}_1^{\bullet}\oplus\mathcal{F}_2^{\bullet}$
Now, my question is about the proof of Lemma 4.5, on page 92, in Huybrecht’s book. This lemma is as follows:
Lemma: Suppose $\mathcal{F}^{\bullet}$ is a simple object in $D^b(X)$ with zero dimensional support. If $Hom(\mathcal{F}^{\bullet},\mathcal{F}[i])=0$ for $i<0$, then
$$\mathcal{F}^{\bullet}\cong k(x)[m]$$
for some closed point $x\in X$ and some integer $m$.
In its proof, to show $\mathcal{F}^{\bullet}$ is concentrated in exactly one closed point, he uses Lemma 3.9, which I stated before, and he get the contradiction if the assumption of Lemma 3.9 holds and then he conclude that we may assume that the support of all cohomology sheaves $H^i(\mathcal{F}^{\bullet})$ consists of the same closed point $x\in X$. My question is that why is this true?

REPLY [3 votes]: Assume on the contrary, that the support of $\mathcal{F}$ consists of multiple closed points. Then by the Lemma you stated, there exist non-trivial $\mathcal{F}_1$ and $\mathcal{F}_2$ with $\mathcal{F} \cong \mathcal{F}_1 \oplus \mathcal{F}_2$. Now we can consider the following morphism:
$$\mathcal{F} \cong \mathcal{F}_1 \oplus \mathcal{F}_2 \twoheadrightarrow \mathcal{F}_1 \hookrightarrow \mathcal{F}_1 \oplus \mathcal{F}_2 \cong \mathcal{F},$$ where the two morphisms in the middle are the projection onto the first factor and the inclusion as first factor. Since this morphism is non-zero and not invertible, its existence contradicts $\mathcal{F}$ being simple. Hence our assumption was wrong, so that the support only consists of a single closed point.