Dualizing Sheaf unique determined For a curve $X$ and a morphism $f: X \to \mathbb{P}^1$ we can define the dualizing sheaf $\omega_f$ by following caracteristic property: $$ f_*(\omega_f) = \underline {Hom} _{\mathcal{O}_{\mathbb{P}^1}}(f_*(\mathcal{O}_X), \omega_{\mathbb{P}^1})$$
where $\omega_{\mathbb{P}^1} = \mathcal{O}_{\mathbb{P}^1}(-2)$.
My question is why is $\omega_f$ by this condition unique determined?
 A: Short answer: because they both represent the functor $\mathcal{F}\to H^n(X,\mathcal{F})'$ from coherent sheaves to $k$-vector spaces, and objects representing functors are unique up to unique isomorphism.
Longer answer: If you have proven that dualizing sheaves fulfill the following properties, the proof is not so hard.
A dualizing sheaf on a proper scheme of dimension $n$ over a field $k$ is a coherent sheaf $\omega$ with a trace morphism $t:H^n(X,\omega)\to k$ such that for all coherent $\mathcal{F}$, the natural pairing $\operatorname{Hom}(\mathcal{F},\omega)\times H^n(X,\mathcal{F})\to H^n(X,\omega)$ followed by $t$ gives an isomorphism $\operatorname{Hom}(\mathcal{F},\omega)\to H^n(X,\mathcal{F})'$, where $'$ denotes dual as a $k$-vector space.
Proof of uniqueness, assuming the above properties:
Suppose $\omega$ and $\omega_1$ are two dualizing sheaves. Then since $\omega_1$ is dualizing, we have an isomorphism $\operatorname{Hom}(\omega,\omega_1)\cong H^n(\omega)'$. So there's a unique morphism $\varphi: \omega\to\omega_1$ corresponding to $t:H^n(X,\omega)\to k$. Similarly, we can get a unique $\psi:\omega_1\to\omega$ corresponding to $t_1:H^n(X,\omega_1)\to k$. The composition of the above maps in both directions must be the identity, so they are both isomorphisms and we are done.
