Chern character of Hom-sheaves I'm reading the book about moduli spaces by Huybrechts and Lehn, and i'm stuck understanding a proof, it is Theorem 6.1.8.:
Given a K3-surface $X$ and a 2-dimensional space $M$, coherent and torsion free sheafs $F$ on $X$ and $E$ on $M\times X$. We have projections $p,q$ from $M\times X$ to $M$ and $X$ resp.
They claim the class $a:=ch(p_{\*}\mathcal{H}om(q^{\*}F,E))$ as an element in $H^{\*}(M,\mathbb{Q})$, where $p_{\*}\mathcal{H}om(q^{\*}F,E)=\sum\limits_{i=0}^2 (-1)^i \mathcal{E}xt^i_p(q^{\*}F,E)$, only depends on the classes of $ch(q^{\*}F)$ and $ch(E)$ as elements in $H^{\*}(M\times X,\mathbb{Q})$, where $\mathcal{E}xt_p^i(q^{\*}F,E)=R^i(p_{\*}\mathcal{H}om(q^{\*}F,E))$.
So using Grothendieck-Riemann-Roch as suggested shows:
$ch(\sum\limits_{i=0}^2 (-1)^i \mathcal{E}xt^i_p(q^{\*}F,E))td(M)=p_{\*}(ch(\mathcal{H}om(q^{\*}F,E)td(M\times X))$
Here i am stuck. Why does this show that $a$ only depends on $ch(E)$ and $ch(q^{\*}F)$. I think one has to show that $ch(\mathcal{H}om(q^{\*}F,E))$ only depends on this classes, but i can't see why.
 A: If you apply ${\mathcal H}om(-,E)$ to a resolution of a sheaf $G$, you obtain a complex, the cohomology of which are ${\mathcal E}xt^i(G,E)$, hence by additivity of the Chern character, the alternating sum of Chern characters of the terms of the complex equals the alternating sum of Chern characters of the Ext sheaves. 
A: This was to long for a comment, so i post this as an answer: Using Sasha's answer i tried my best, and here are my computations. Feel free to report any mistakes. 
Take a locally free resolution $G_{\*} \rightarrow q^{\*}F$. Then we have $\mathcal{E}xt_p^i(q^{\*}F,E)=h^i(\mathcal{H}om_p(G_{\*},E))$, where $h^i = ker(d_i)/im(d_{i-1})$. Now:
$ch(\sum\limits_{i=0}^2(-1)^i \mathcal{E}xt_p^i(q^{\*}F,E))=ch(\sum\limits_{i=0}^2 (-1)^i h^i(\mathcal{H}om_p(G_{\*},E)))$ 
$= ch(ker(d_0))-ch(ker(d_1)/im(d_0))+ch(coker(d_1))$
$=ch(ker(d_0))+ch(im(d_0))-ch(ker(d_1))-ch(im(d_1))+ch(\mathcal{H}om_p(G_2,E))$
$ch(ker(d_0)\oplus im(d_0))-ch(ker(d_1)\oplus im(d_1))+ch(\mathcal{H}om_p(G_2,E))$
$ch(\mathcal{H}om_p(G_0,E))-ch(\mathcal{H}om_p(G_1,E))+ch(\mathcal{H}om_p(G_2,E))$ (1)
Now one uses Grothendieck-Riemann-Roch:
$ch(\mathcal{H}om_p(G_i,E))=ch(p_{\*}\mathcal{H}om(G_i,E))=p_{\*}(ch(\mathcal{H}om(G_i,E))td(X))$
Now the $G_i$ are locally free, so $ch(\mathcal{H}om(G_i,E))=ch(G_i^{\*})ch(E)$ and (1) gives:
$p_{\*}(ch(G_0^{\*})ch(E)td(X))-p_{\*}(ch(G_1^{\*})ch(E)td(X))+p_{\*}(ch(G_2^{\*})ch(E)td(X))$
which is $p_{\*}((\sum\limits_{i=0}^2 (-1)^i ch(G_i^{\*}))ch(E)td(X))$, but since $G_{\*}$ was a resolution  of $q^{\*}F$ this is
$p_{\*}(ch((q^{\*}F)^{\*})ch(E)td(X))$
So we have: $ch(\sum\limits_{i=0}^2(-1)^i \mathcal{E}xt_p^i(q^{\*}F,E))=p_{\*}(ch((q^{\*}F)^{\*})ch(E)td(X))$
So i think my assumption, that $ch(\mathcal{H}om(M,N))=ch(M^{\*})ch(N)$ is always true, is wrong?
