# Universal mapping property for projective schemes and globalising it to projective bundles

Let $$Y = \text{Spec}A$$ be a noetherian affine scheme. Let $$S$$ be a graded $$A$$-algebra which is finitely generated by $$S_{1}$$ as an $$S_{0}$$ algebra. In other words, $$S$$ looks like $$S = A[x_{0}, x_{1}, \ldots , x_{n}]/I$$ for some homogenous ideal $$I$$. Let $$\mathcal{F}$$ be the sheaf of algebras $$\widetilde{S}$$ on $$\text{Spec}A$$. I am trying to understand the projective space $$\text{Proj}_{Y} \mathcal{F}$$ via universal properties. My understanding so far is this: The scheme $$\text{Proj}_{Y} \mathcal{F}$$ is defined to be the $$A$$-scheme such for any other $$A$$-scheme $$\mu: T \rightarrow \text{Spec}A$$, to give an $$A$$-morphism from $$T$$ to $$\text{Proj}_{Y} \mathcal{F}$$ is to give a line bundle quotient $$\mu^{*} \mathcal{F} \stackrel{q}{\longrightarrow} \mathcal{L} \longrightarrow 0.$$ Another universal property is to say that to give an $$A$$-morphism from $$T$$ to $$\text{Proj}_{Y}\mathcal{F}$$ is to give a line bundle $$\mathcal{L}$$ on $$T$$ along with a morphism of $$A$$-modules $$\phi: S_{1} \rightarrow \Gamma(T, \mathcal{L})$$ so that the image of $$\phi$$ generates $$\mathcal{L}$$.

My question is, why are these equivalent? More precisely,

Why is giving a surjection of sheaves $$\mu^{*} \mathcal{F} {\rightarrow} \mathcal{L} \rightarrow 0$$ equivalent to giving a morphism of $$A$$-modules $$\phi: S_{1} \rightarrow \Gamma(T, \mathcal{L})$$ whose image is a family of global sections of $$\mathcal{L}$$ which generates $$\mathcal{L}$$?

I am assuming the answer involves taking the given surjection $$q: \mu^{*} \mathcal{F} \rightarrow \mathcal{L}$$ and corresponding the adjunct morphism $$q': \mathcal{F} \rightarrow \mu_{*} \mathcal{L}$$. Taking global sections give us a morphism, $$\Gamma(\text{Spec}A, \mathcal{F}) \longrightarrow \Gamma(T, \mathcal{L}).$$ Somehow I need to consider only the degree $$1$$ part of this to obtain a morphism of $$A$$-modules and show that it generates $$\mathcal{L}$$.