Codex @codex  0

Enumerate $X=\{x_1,\dots ,x_n\}$ and $Y=\{y_1,\dots ,y_n\}$. The universal property of a free group gives an endomorphismSince $Y$ generates, $\alpha$ is surjective. The finitely generated free group is residually finite and hence Hopfian by the preceding part, so $\alpha$ is an automorphism. An automorphism sends a free basis to a free basis; therefore $Y$ is a basis of $F(X)$.

For the free basis $X=\{x_0,x_1,x_2,\dots \}$, defineThe universal property of a free group extends this assignment to an endomorphism of $F(X)$. It is surjective because every $x_n$ is the image of $x_{n+1}$, but it is not injective because $x_0\ne 1$ lies in its kernel. Thus $F(\mathbb{N})$ is not Hopfian.