Why is it that if $i:Z\to X$ is an immersion of schemes, then $Z\to X\setminus\partial Z$ is a closed immersion of schemes?

Question

There is such a description in the Stacks Project, considering $i: Z \rightarrow X$ is a immersion, then $i^{'}: Z \rightarrow X \setminus \partial Z$ is a closed immersion. I don't understand why there is such a conclusion. Since $i = f \circ g$, where $f$ is open immersion and $g$ is closed immersion, does this mean $g = i^{'}$?

Answer 1

An immersion is a closed immersion iff it has closed image (ref). $X\setminus \partial Z$ is an open subset of $X$ containing $i(Z)$ so that $i(Z)$ is closed in $X\setminus \partial Z$ (this is purely topological; this is also the largest such open subset), hence you may always factorize any given immersion as $i':Z\to X\setminus \partial Z$ followed by the open immersion of $X\setminus \partial Z \to X$. This means one can always choose $f,g$ so that $g=i'$, but there may be many other choices of factorization of $i$ in to a closed immersion followed by an open immersion.