The determinant line bundle of a coherent sheaf $\mathcal{F}$ on an $n$-dimensional (smooth) analytic space is defined as
\begin{equation}
\det \mathcal{F} := \bigotimes_i^n (\det \mathcal{E}_i)^{⊗ (-1)^i}
\end{equation}
where $\mathcal{E}_\bullet \to F$ is a locally free resolution of $\mathcal{F}$ (which we can take to have length at most $n$). It can be shown that this is independent of the resolution taken, and that if $\mathcal{F}$ is torsion-free then
\begin{equation}
\det \mathcal{F} \cong \left(\bigwedge^{\operatorname{rk} \mathcal{F}} \mathcal{F}\right)^{**}
\end{equation}
. Since $\bigwedge^k$ and double duals are both functorial we see that a morphis $\mathcal{F} \to \mathcal{F}$ of **torsion-free* shaves **of the same rank** induces a morphism between their determinant line bundles. Can we say the same thing about any two coherent sheaves of the same rank?

**My thoughts so far:** It seems like the obvious way to do this would be to take free resolutions $\mathcal{E}_\bullet, \mathcal{E}'_\bullet$ of $\mathcal{F}, \mathcal{F}'$, which gives a map $f_\bullet: \mathcal{E}_\bullet \to \mathcal{E}'_\bullet$ and to just take the alternating tensor product of the maps induced by $f_i$ from $\det \mathcal{E}_i \to \det \mathcal{E}'_i$. However those maps don't exist because $\det$ is only functorial on vector bundles of the same rank?