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Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi^\ast:M/M^2 \to N/N^2$. Here both $M/M^2$ and $N/N^2$ are $n$-dimensional $k$-vector spaces, and $\Phi^\ast$ is clearly an isomorphism. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have $\Phi^\ast(f) = 0$, a contradiction. Thus $\Phi$ is injective.