I'm reading the proof of the following statement in Milne's book "Algebraic geometry".
Chow's Lemma: Let $V$ be a complete irreducible variety. There exists a projective variety $V'$ and a surjective map $f:V'\to V$ such that for some dense open subset $U\subseteq V$ the map $f:f^{-1}(U)\to U$ is an isomorphism.
At page 168 of the book the author says
"We first show that $g:V'\to P$ is an immersion. As this is a local condition, it suffices to find open subsets $V_i\subseteq P$ such that $V'\subseteq\bigcup q^{-1}(V)$ and each map $g:V'\cap q^{-1}(V_i)\to V_i$ is an immersion."
How is this enough? Immersions are injective so we are saying that a locally injective map is injective which is not true. Am I missing something?
Immersions are local on the target - that is, if there is a map $f:X\to S$ and an open covering $\{U_i\}_{i\in I}$ of $S$ so that $f^{-1}(U_i)\to U_i$ is an immersion for all $i\in I$, then $f$ is an immersion. This works because if $x_1,x_2$ are two points mapping to $s\in S$, then for any open $U\subset S$ containing $s$, the map $f^{-1}(U)\to U$ won't be injective.
In your complaint, you may be thinking of a locally injective map, where injectivity is local on the source - i.e. for each $x\in X$, there's an open neighborhood $U_x$ of $x$ so that $f|_{U_x}: U_x\to S$ is injective. That's indeed not necessarily an injective map: $pt \sqcup pt \to pt$ is of course an example. But that's not the concept being discussed here.