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Let $f:X \rightarrow Y$ be a map of $m$-dimensional simplicial spaces (which means that all simplices above dimension $m$ are degenerate). Recall, that $f$ is a natural transformation of functors from $\Delta$ to spaces. I want to call such a map proper, if each $f_n:X_n\rightarrow Y_n$ is proper.

So the question is, whether $f$ is proper if and only if $|f|$ is proper.

The finite dimensionality is required, as the following example shows: Take $X$ to be any simplicial space with a finite, positive number of nondegenerate simplices in each dimension. Then the map $f:X\rightarrow pt$ is proper (in the notation from above), but $|X|$ is not compact and hence $|f|$ is not proper.

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Can you clarify what you mean by "pointwise proper"? – Chad Groft Mar 25 '10 at 2:39
Probably levelwise proper, i.e., $f_n : X_n \to Y_n$ is proper for each $n \ge 0$. – Reid Barton Mar 25 '10 at 3:42
Didn't you forget to add some tag like "category theory"? I'm afraid that the only answer from a topologist can be: "How on Earth can a simplex of a simplicial space (=simplicial complex) be degenerate?" – Sergei Ivanov Mar 25 '10 at 19:20
Although I think, that every topologist is familiar to the concept of simplicial objects in a category, I added the tag :). – HenrikRüping Mar 25 '10 at 21:58

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