# Is the determinant line bundle of a coherent sheaf functorial (between sheaves of the same rank)?

The determinant line bundle of a coherent sheaf $$\mathcal{F}$$ on an $$n$$-dimensional (smooth) analytic space is defined as $$$$\det \mathcal{F} := \bigotimes_i^n (\det \mathcal{E}_i)^{⊗ (-1)^i}$$$$ where $$\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}$$ is torsion-free then $$$$\det \mathcal{F} \cong \left(\bigwedge^{\operatorname{rk} \mathcal{F}} \mathcal{F}\right)^{**}$$$$ . Since $$\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?

Working on projective space, consider a composition $$\mathcal O \to \mathcal O \oplus \mathcal O/\mathcal O(-1) \to \mathcal O$$ where $$\mathcal O/\mathcal O(-1)$$ is the constant sheaf on a hyperplane, and the maps are just defined to be the identity on $$\mathcal O$$ and $$0$$ on $$\mathcal O/\mathcal O(-1)$$.
But taking determinants, we get $$\mathcal O \to \mathcal O(1) \to \mathcal O$$ and since every map $$\mathcal O(1) \to \mathcal O$$ vanishes, there is no composition of maps $$\mathcal O \to \mathcal O(1)$$ and $$\mathcal O(1) \to \mathcal O$$ that gives the identity.