Suppose I have a morphism of Noetherian schemes over a field $k$ (if one needs this then assume $k$ is algebraically closed) $f:C'\rightarrow S$ which is finite with geometrically connected and reduced fibers. Is $f$ an isomorphism?
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No: $\text{Spec} k[\epsilon]/\langle \epsilon \rangle$ mapping to $\text{Spec} k[\epsilon]/\langle \epsilon^2 \rangle$. 


Though any closed embedding will give a counter example, it is true if you ask whether it is an isomorphism to the (scheme theoretic) image. In other words, the map is an isomorphism from $C'\to f(C')$ where $f(C')$ is thought of as the scheme theoretic image (which makes sense since the map is assumed to be finite), with $k$ algebraically closed. 


Or, perhaps more to the point, the map from the normalization to a cusp on a curve. Note that such a map is an isomorphism on points without being an isomorphism of schemes. 

