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.

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 $Y$. 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 


Though it is not true in general, it is true whenever $Y$ is separated. The map $f$ from $X$ to $Y$ factors as a composition $$ X \stackrel{f'}{\rightarrow} X \times Y \stackrel{f''}{\rightarrow} Y$$ The map $f''$ is a pullback of the projection map from $X$ to a point (or Spec $\mathbb{Z}$, or whatever base you are working over), and therefore affine. The map $f'$ is a pullback of the diagonal $Y \rightarrow Y \times Y$, and therefore a closed immersion if $Y$ is separated (and in particular affine). 

