We prove this for the total Chern class; the other two cases follow by the same arguments using Lemma \ref{lemma-chern-character-additive} instead of Lemma \ref{lemma-additivity-chern-classes}. \medskip\noindent Suppose we have a second finite complex $\mathcal{F}^\bullet$ of finite locally free $\mathcal{O}_X$-modules representing $E$. Let $g : Y \to X$ be a morphism locally of finite type with $Y$ integral. By Lemma \ref{lemma-bivariant-zero} it suffices to show that with $c(g^*\mathcal{E}^\bullet) \cap [Y]$ is the same as $c(g^*\mathcal{F}^\bullet) \cap [Y]$ and it even suffices to prove this after replacing $Y$ by an integral scheme proper and birational over $Y$. By More on Flatness, Lemma \ref{flat-lemma-blowup-complex-integral} we may assume that $H^i(Lg^*E)$ is perfect of tor dimension $\leq 1$ for all $i \in \mathbf{Z}$. This reduces us to the case discussed in the next paragraph. \medskip\noindent Assume $X$ is integral and $H^i(E)$ is a perfect $\mathcal{O}_X$-module of tor dimension $\leq 1$ for all $i \in \mathbf{Z}$. We have to show that $c(\mathcal{E}^\bullet) \cap [X]$ is the same as $c(\mathcal{F}^\bullet) \cap [X]$. Denote $d_\mathcal{E}^i : \mathcal{E}^i \to \mathcal{E}^{i + 1}$ and $d_\mathcal{F}^i : \mathcal{F}^i \to \mathcal{F}^{i + 1}$ the differentials of our complexes. By More on Flatness, Remark \ref{flat-remark-when-you-have-a-complex} we know that $\Im(d_\mathcal{E}^i)$, $\Ker(d_\mathcal{E}^i)$, $\Im(d_\mathcal{F}^i)$, and $\Ker(d_\mathcal{F}^i)$ are finite locally free $\mathcal{O}_X$-modules for all $i$. By additivity (Lemma \ref{lemma-additivity-chern-classes}) we see that $$ c(\mathcal{E}^\bullet) = \prod\nolimits_i c(\Ker(d_\mathcal{E}^i))^{(-1)^i} c(\Im(d_\mathcal{E}^i))^{(-1)^i} $$ and similarly for $\mathcal{F}^\bullet$. Since we have the short exact sequences $$ 0 \to \Im(d_\mathcal{E}^i) \to \Ker(d_\mathcal{E}^i) \to H^i(E) \to 0 \quad\text{and}\quad 0 \to \Im(d_\mathcal{F}^i) \to \Ker(d_\mathcal{F}^i) \to H^i(E) \to 0 $$ we reduce to the problem stated and solved in the next paragraph. \medskip\noindent Assume $X$ is integral and we have two short exact sequences $$ 0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{Q} \to 0 \quad\text{and}\quad 0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{Q} \to 0 $$ with $\mathcal{E}$, $\mathcal{E}'$, $\mathcal{F}$, $\mathcal{F}'$ finite locally free. Problem: show that $c(\mathcal{E})c(\mathcal{E}')^{-1} \cap [X] = c(\mathcal{F})c(\mathcal{F}')^{-1} \cap [X]$. To do this, consider the short exact sequence $$ 0 \to \mathcal{G} \to \mathcal{E} \oplus \mathcal{F} \to \mathcal{Q} \to 0 $$ defining $\mathcal{G}$. Since $\mathcal{Q}$ has tor dimension $\leq 1$ we see that $\mathcal{G}$ is finite locally free. A diagram chase shows that the kernel of the surjection $\mathcal{G} \to \mathcal{F}$ maps isomorphically to $\mathcal{E}'$ in $\mathcal{E}$ and the kernel of the surjection $\mathcal{G} \to \mathcal{E}$ maps isomorphically to $\mathcal{F}'$ in $\mathcal{F}$. (Working affine locally this follows from or is equivalent to Schanuel's lemma, see Algebra, Lemma \ref{algebra-lemma-Schanuel}.) We conclude that $$ c(\mathcal{E})c(\mathcal{F}') = c(\mathcal{G}) = c(\mathcal{F})c(\mathcal{E}') $$ as desired.