Cover of a Grassmannian by an open set I am reading this document here and in exercise 1, the author asks to show the Grassmannian $G(r,d)$ in a $d$ dimensional vector space $V$ has dimension $r(d-r)$ as follows. For each $W \in G(r,d)$ choose $V_W$ of dimension $d-r$ that intersects $W$ trivially, and show one has a bijection
$$\{ \text{Subspaces of dimension $r$ that intersect $V_W$ trivially}\} \leftrightarrow \operatorname{Hom}(W,V_W).$$
Now I can set up a bijection as follows. For each $U$ on the left, we have $U \oplus V_W = V$. Thus any $w \in W$ can be written uniquely as $u + v_w$ for $u\in U$ and $v_w \in V_W$ and so we obtain a linear map $T : W \to V_W$ by sending $w$ to this $v_w$. Conversely for any linear map $T \in \text{Hom}(W,V_W)$ we have a subspace on the left consisting of vectors $w + T(w)$ for $w \in W$. 


My question is: The author has not defined why the set on the left hand side above is an open set. Presumably he meant that if we look in $\Bbb{P}(V)$, the set of all linear varieties of dimension $r -1$ that do intersect  a given linear variety of dimension $d-r - 1$ is closed (in the Zariski topology), but why is this so?



As the author says, he has not defined a topology on the Grassmannian yet. The point is that the covering he describes can be taken as defining the topology: you take this as your atlas, i.e. by glueing. You only need to check that the transition maps are smooth (with respect to the natural topology on Hom-sets).
I will not spoil the exercise for you, but if you compute the transition maps you will see they look like 'fractional linear transformations'.
(Also see this question.)
