To put this question in precise language, let $X$ be an affine scheme, and $Y$ be an arbitrary scheme, and $f : X \rightarrow Y$ a morphism from $X$ to $Y$. Does it follow that $f$ is an affine morphism of schemes? While all cases are interesting, a counterexample that has both $X$ and $Y$ noetherian would be nice.



Are morphisms from affine schemes to arbitrary schemes affine morphisms?
No, here is an example of a morphism $f:X\to Y$ which is not affine although $X$ is affine.


Take $X=\mathbb A^2\_k$, the affine plane over the field $k$ and for $Y$ the notorious plane with origin doubled: $Y=Y\_1\cup Y\_2$ with $Y\_i\simeq \mathbb A^2\_k$ open in $Y$ and $Y\setminus Y\_i= \lbrace O\_i\rbrace$, a closed rational point of $Y$.  

We take for $f:X\to Y$ the map sending $X$ isomorphically to $Y\_1$ in the obvious way. 


Then, although the scheme $X$ is affine, the morphism $f$ is not affine because the inverse image $f^{-1}(Y\_2)$of the affine open subscheme $Y\_2\subset Y$ is  

 $X \setminus \lbrace 0 \rbrace=\mathbb A^2\_k \setminus \lbrace 0 \rbrace$, the affine plane with origin deleted, well known not to be affine.


