# Show that there is $\tilde\varphi$ which makes a diagram involving reduced finitely generated $k$-algebras commutative

Let $k$ be an algebraically closed field, $X\subseteq \mathbb{A}^m(k)$, $Y\subseteq \mathbb{A}^n(k)$ affine algebraic sets and $\varphi:\Gamma(Y)\to\Gamma(X)$ be a morphism of reduced finitely generated $k$-algebras.

I want to prove that there is a $\tilde\varphi:k[T_1',\dots,T_n']\to k[T_1,\dots,T_m]$ such that $$k[T_1',\dots,T_n'] \xrightarrow{\tilde\varphi} k[T_1,\dots,T_m] \to \Gamma(X)$$ $$k[T_1',\dots,T_n'] \to \Gamma(Y) \xrightarrow{\varphi} \Gamma(X)$$ commutes. Can you help me?

Let $\pi':k[T_1',\dots,T_n']\to\Gamma(Y)$ and $\pi:k[T_1,\dots,T_m]\to\Gamma(X)$ be the canonical maps.
It suffices to define $\tilde\varphi$ for the $T_i'$. In order to make the diagram commutative just pick an arbitrary $f_i\in \pi^{-1}(\varphi(\pi(T_i')))$ and assign $T_i'\mapsto f_i$ for all $i$ with $1\le i \le n$.