Let $k$ be an algebraically closed field and let $V$ denote a $k$-variety. Denote by $t(V)$, the set of all irreducible, closed sets of $V$, where irreducible means that $S$ cannot be written as a union of two closed subsets of $S$.
We show that $t(V)$ is isomorphic as an affine scheme to $\operatorname{spec} A$ where $A$ is the coordinate ring of $V$. The map of topological spaces, $\alpha$ is given by mapping $p\in V$ to $\{p\}^-$. This induces the map on sheaves such that $\mathcal{O}_{t(V)}(U) = \mathcal{O}_{V}(\alpha^{-1}U)$ for $U\in t(V)$. We can show that $\alpha$ is a bijection on open sets of $V$ and $t(V)$.
Also, there exists a bijection from points of $V$ onto closed points of $\operatorname{spec} A$ given by $\beta(p) = \mathfrak{m_p} = \{f\in\mathcal{}O_V\mid f(p) = 0\}$.
For varieties $V,W$ we have $T:\hom(V,W) \rightarrow \hom(t(V), t(W))$ mapping $f:V\rightarrow W$ to $T(F):t(V)\rightarrow t(W)$ such that $$T(f)(C) = f(C)^-$$ where $S^-$ denotes the closure in $V$.
We are also given that closed points of $t(V)$ map to closed points of $t(W)$ and if $p\in V$ is closed, then its image in $t(V)$ is $\{p\}$.
I looked at all answers involving this exercise but I can't figure out why $T$ is injective.
My attempt
For some $f,g:V\rightarrow W$, we are given $T(f) = T(g)$, thus for any element $C\in t(V)$, we have $$f(C)^- = g(C)^-.$$
Is there a reason why $f(C)$ is closed in $W$? (It is evidently a closed point of $t(W)$).
I even assumed $C = \{p\}^-$ for some point $p\in V$ but that gives nothing.
Suppose $f,g:V\to W$ are two distinct maps, i.e. there is some $v\in V$ so that $f(v)\neq g(v)$. Now look at $T(f)(\{v\})$ and $T(g)(\{v\})$: these are just $\{f(v)\}$ and $\{g(v)\}$, which aren't equal and thus $T(f)\neq T(g)$.