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, iff it if each $f_n:X_n\rightarrow Y_n$ is pointwise 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.

Let $f:X\rightarrow f:X \rightarrow Y$ 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, iff it is pointwise 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.

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, iff it is pointwise 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.