I apologize in advance if my question is too elementary for MO.

It is a well known fact that the linear algebraic group $G = \mathsf{SO}_n$ is connected, and there exist a few different proofs of this fact. One proof goes by showing that $G$ is generated by unipotent elements, and invoking the theorem that every linear algebraic group with this property is connected.

My question is about a different, more direct proof, involving the Cayley transform
$$ A \mapsto (I_n + A)^{-1} (I_n - A) , $$
which maps every matrix $A \in G(k)$ for which $I_n + A$ is invertible (let's write $W$ for the set of such matrices), to a skew-symmetric matrix, and in fact, this map defines an isomorphism of varieties between the non-empty open subset $W$ of $G$, and an open subset of the irreducible variety $V$ of skew-symmetric matrices.

But how does one now conclude that $\mathsf{SO}_n$ is connected? In particular, why is the closure of $W$ equal to $G$? Of course the complement of $W$ is given by the polynomial equation $\det(I_n + A) = 0$, but how is this situation different from, for instance, the fact that $\mathsf{SO}_n$ is the subvariety of $\mathsf{O}_n$ defined by the polynomial equation $\det(A) - 1 = 0$ (or $\det(A) + 1 \neq 0$) whereas $\mathsf{O}_n$ is not connected?

What subtlety am I missing?

**added:**

 1. I'm assuming $\operatorname{char}(k) \neq 2$.
 2. The underlying topology is the Zariski toplogy.
 3. The proof I'm mentioning in the second paragraph is only valid in characteristic $0$.
 4. It seems that the question is not too elementary...

**added:** I'll add a few more sentences to make clear what I mean by connected in the Zariski topology.

I'm considering $G$ as a group functor (i.e. a functor from the category of commutative $k$-algebras to the category of groups). Then $G$ is connected if and only if its coordinate algebra $k[G] = \mathcal{O}(G)$ has no non-trivial idempotents. In particular, in order to show that $G$ is connected, we may assume w.l.o.g. that $k$ is algebraically closed. [In fact, since $G$ is smooth (since $\operatorname{char}(k) \neq 2$), we may instead consider the group of $\bar{k}$-rational points $G(\bar{k})$, which brings us back to a more classical situation.]