Can a non-surjective finite morphism exist?
Solution
1. Let\[ i:\{0\}\longrightarrow\AA^1_k \]be the inclusion of the origin. Then \(i\) is finite.
The comorphism is the quotient map \[ i^*:k[t]\longrightarrow k, \qquad t\longmapsto0, \] or equivalently \[ k[t]\longrightarrow k[t]/(t). \] As a \(k[t]\)-module, \[ k[t]/(t) \] is generated by the class of \(1\). Hence it is a finite \(k[t]\)-module, so \(i\) is finite.
2. The finite morphism \(i\) is not surjective.
Its image is \[ i(\{0\})=\{0\}\subsetneq\AA^1_k. \] Thus a finite morphism need not be surjective when its comorphism is not injective. In this example \[ \dim\{0\}=0<1=\dim\AA^1, \] so the hypothesis \(\dim X=\dim Y\) of the surjectivity criterion P-AGXVARFINCLOSED fails.
3. Q.E.D.
Steps <1>1–<1>2 give a non-surjective finite morphism, so the answer is \(\boxed{\text{yes}}\).