how to show f is bijective?

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Suppose that the set-mapping $f:X\rightarrow Y$ of one-dimensional domains of $\mathbb{C}$ induces an isomorphism $f^0:\mathcal{O}(Y)\rightarrow \mathcal{O}(X)$ defined by $g\mapsto g\circ f$ of topological algebras.Show that $f$ is bijective.\ My idea is to suppose $f$ is not bijective,dividing this problem into two cases,i.e. $f$ is not injective or $f$ is not surjective,then I want to conduct a contradiction that $f^0$ cannot be an isomorphism in any two cases.But I can't find where the contradiction lies.Can you help me?Thanks in advance!

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If $f$ is not injective, say $f(z_1) = f(z_2) = w_0 \in Y$, let $h(w) = w-w_0$. Then $h$ is prime in $\mathcal{O}(Y)$. What about $f^0(h)$?

If $f$ is not surjective, let $w_0 \in Y \setminus f(X)$, and $h(w) = w-w_0$. What can you say about $f^0(h)$ that distinguishes it from $h$ algebraically?