Given two torsion free coherent sheaves $M$ and $N$ wit $rk(M)=rk(N)=r$ on an smooth projective surface $S$, by definition $det(M):=\Lambda^r(M)^{\\*\\*}$.


Is the following criterion correct?


$M\cong N$ $\Leftrightarrow$ $M \hookrightarrow N$ and $c\_i(M)=c\_i(N)$ for $i=0,1,2$


One only has to look at "$\Leftarrow$":


So we have $0\rightarrow M\rightarrow N\rightarrow Q \rightarrow0$. Because of $c\_i(M)=c\_i(N)$ we see that $c\_1(Q)=0$ and $c\_2(Q)=0$. Since $M$ and $N$ have the same rank $codim(supp(Q))\geq 1$. We also have an induced map $det(M)\hookrightarrow det(N)$ of line bundles, i.e. $det(N)\cong det(M)\otimes O\_S(D)$ for some effective divisor $D$. Now $det(Q)\cong det(N)\otimes det(M)^{-1}\cong O\_S(D)$. 


So by definition $c\_1(Q)=c\_1(det(Q))=D$, but $c\_1(Q)=0$, so $D$ is effective and 0, i.e. $Q$ has no support in codimension 1, so $codim(supp(Q))\geq 2$. 


So $Q$ is an Artinian sheaf, and for those one has $c\_2(Q)=-dim(H^0(S,Q))$. Since $c\_2(Q)=0$ we have $H^0(S,Q)=0$, but $H^0(S,Q)=\bigoplus\limits\_{s\in supp(Q)} Q\_s$. So $Q\_s=0$ for all $s\in supp(Q)$, i.e. $Q=0$. So we have $M\cong N$.



Equality of chern classes and isomorphism
That's correct. A slightly shorter argument is: if $\mathcal{Q}$ has support in codimension $d$, then its Chern character $\mathrm{ch}\_d(\mathcal{Q})$ is non-zero and effective. So a sheaf is trivial if and onlfy if $\mathrm{ch} = 0$, which is true if and only if the rank and the Chern classes vanish. In particular, the result holds for any dimension.


