Prove that $\Gamma$ is an equivalence of categories between the category of affine algebraic sets and the category of reduced $k-$ algebras

I am self studying Algebraic Geometry from the Daniel Perrin's textbook.

$k$ is an algebraically closed field. $\Gamma(V)$ is image of $r: k[X_1,...,X_n]\to F(V,k)$ and $V,W$ is affine algebraic set. $\theta : \Gamma(W) \to \Gamma(V)$ be a homomorphism of $k-$algebras and $\phi: V\to k^m$ is the map whose co-ordinates are $\phi_i$. If $\phi:V\to W$be a morphism such that $V\subset k^n $ and $W\subset k^m$ written in the form $\phi= (\phi_1,...,\phi_m)$ , where $ \phi_i \in \Gamma(V)$. Set $\phi_i =\theta(\eta_i)\in \Gamma(V)$ Here $ \eta_i$ is the $i$ th coordinate function of $W$, which is the image of the variable $y_i$ in $\Gamma(W)$

I have a question in the proof of Theorem 6.13 on Page 23 which is:

Assume that $k$ is algebraically closed. The functor $\Gamma$ is then an equivalence of categories between the category of affine algebraic sets with regular maps and the category of reduced $k-$algebras of finite type with homomorphisms of $k-$algebras.

Proof: Injectivity has already been proved in Prop $6.7$ and is clear to me.

For surjectivity :$A$ can be written as $A\cong k[x_1,...,x_n]/I$ as $A$ is of the finite type and $A$ is reduced implies that ideal $I$ is radical . Set $V=V(I)$ , we have I(V)= rac(I)=I by nullstellansatz ( all of this is clear to me except:)

How does all this implies that $A\cong \Gamma(V)$?

Can you please help me understand this?