I'm currently reading a paper by Nakajima (Quiver Varieties and Tensor Products), and I'm having a hard time understanding a very specific step in his proof of Lemma 3.2. Essentially, we have two (quasi-projective) varieties, say $X$ and $Y$, that we would like to show are isomorphic. The proof uses the following argument:

(1) Construct a bijective morphism $f:X \to Y$.

(2) Show that $\mathrm{d}f: T_x(X) \to T_{f(x)}(Y)$ is an isomorphism for all $x \in X$, where $T_x(X)$ is the tangent space of $X$ at $x$.

From (1) and (2), he concludes that $f$ is an isomorphism. My question is, how can one conclude that f is an isomorphism from (1) and (2)?

In doing a little research, I found this topic. To summarize Theorem 14.9 and Corollary 14.10 in Joe Harris' Algebraic Geometry: A First Course:

If $f:X \to Y$ is a morphism of varieties and either $f$ is finite or $X$ and $Y$ are projective, then $f$ is an isomorphism if and only if $f$ is bijective and $\mathrm{d} f: T_x(X) \to T_{f(x)}(Y)$ is injective for all $x \in X$.

In the context of Lemma 3.2 in Nakajima's paper, neither variety is projective, which leaves two possibilities. Either the morphism constructed in the proof is in fact finite (but this, to me at least, is not at all obvious), or he is using some other fact that allows him to conclude that $f$ is an isomorphism given (1) and (2) (potentially something about the symplectic form on the quiver varities). Does anyone have any insights as to what facts Nakajima might be using in his proof?