# Can any radiciel morphism be presented as the composition of a universal homeomorphism with an immersion?

Let $f:X\to Y$ be a radiciel (=universally injective) morphism, where $X$ is a regular connected scheme. Can it be presented as the composition of a universal homeomorphism with an immersion? This seems to be equivalent to: the images of points of $X$ yield a subscheme of $Y$.

As in my previous questions, I am interested in excellent schemes of finite Krull dimension. If the answer to my question is 'no', then I would like to know whether there exists a 'short' characterization of the compositions of universal homeomorphisms with immersions.

Upd. As the comments show, the statement is wrong for trivial reasons if we don't assume that $X$ is connected.

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There are lots of monomorphisms of schemes (even of finite type) which are not of this form: take $Y=\mathbb{A}^1_k$ ($k$ a field) and take for $X$ the disjoint union of $\mathbb{G}_{m,k}$ and the origin. You need to assume, at least, that $f$ is a homeomorphism on its image, plus something to ensure that the image is locally closed (think of the inclusion of a generic point). – Laurent Moret-Bailly Mar 22 '11 at 15:51
I just saw Laurent's comment. Since he makes exactly the same points as my answer, and was posted earlier, I deleted my answer. – Angelo Mar 22 '11 at 16:24
Thank you! Actually, I am mostly interested in the case when $X$ is regular and connnected (see the update). Does this help? – Mikhail Bondarko Mar 22 '11 at 19:15
Well, take $Y$ to be an irreducible curve with one node, and $X=$ the normalization minus one of the points above the node. – Laurent Moret-Bailly Mar 22 '11 at 20:03