I'm reading Hartshorne's *Deformation Theory* book. In exercise 5.9(b), he claims:


Let $C$ be a reduced projective locally complete intersection curve embedded inside a smooth surface $X$ (over some alg. closed field $k$), as:
$$i : C\hookrightarrow X$$
Tensoring the exact sequence
$$0\rightarrow T\_{C/k}\rightarrow i^\*T\_{X/k}\rightarrow N\_{C/X}\rightarrow T^1\_{C/k}\rightarrow 0$$
with $\omega\_{X/k}$ (the dualizing sheaf of $X$, ie $\omega\_{X/k} = \det\Omega\_{X/k}$), we get a sequence
$$0\rightarrow T\_{C/k}\otimes\omega\_{X/k}\rightarrow i^\*\Omega\_{X/k}\rightarrow \omega\_{C/k}\rightarrow T^1\_{C/k}\otimes\omega\_{X/k}\rightarrow 0$$
here $T = \Omega^\vee$ denotes the relative tangent sheaf, and $N$ the relative normal sheaf. The extremal terms of the sequence obviously are obtained by tensoring with $\omega\_{X/k}$. The third term is $\omega\_{C/k}$ by essentially the definition of the dualizing sheaf of an LCI scheme, so my question concerns the second term - 


Why is $i^\*T\_{X/k}\otimes\omega\_{X/k}\cong i^\*\Omega\_{X/k}$? Here, I assume the left hand side should be viewed as 
$$i^\*T\_{X/k}\otimes\omega\_{X/k} := T\_{X/k}\otimes\_{\mathcal{O}\_X}\omega\_{X/k}\otimes\_{\mathcal{O}\_X}\mathcal{O}\_C = i^\*\left(T\_{X/k}\otimes\omega\_{X/k}\right)$$



For a LCI curve $C$ embedded inside a smooth surface $i : C\rightarrow X/k$, is $i^*T_X\otimes\omega_X = i^*\Omega_X$?
Let $\mathscr F$ be a rank $n$ locally free sheaf on a ringed space $X$. Then the wedge product map
$$\bigwedge^r \mathscr F \otimes \bigwedge^{n-r} \mathscr F \to \bigwedge^n \mathscr F$$
is a perfect pairing. Indeed, it suffices to check this locally, where one does this by choosing a basis $x\_i$ for $\mathscr F$. Indeed, given $I = \{i\_1 < \ldots < i\_r\}$ with complement $I^\mathsf{c} = \{i\_{r+1} < \ldots < i\_n\}$, write $x\_I = x\_{i\_1} \wedge \ldots \wedge x\_{i\_r}$, and write $\sigma\_I \colon \{1,\ldots,n\} \to \{1,\ldots,n\}$ for the permutation $k \mapsto i\_k$. Then the dual basis of $\{x\_I\}\_I$ is $\{\operatorname{sgn}(\sigma\_I)\ x\_{I^{\mathsf{c}}}\}\_I$.


Applying this to $\Omega^1\_{X/k}$ for $X$ smooth projective of dimension $n$, we get a perfect pairing
$$\Omega^1\_{X/k} \otimes \Omega^{n-1}\_{X/k} \to \omega\_{X/k},$$
realising $\Omega^{n-1}\_{X/k}$ as $\mathscr Hom(\Omega^1\_{X/k},\omega\_{X/k}) = T\_{X/k} \otimes \omega\_{X/k}$.


