On showing that $U_E$ is open in $\mathcal {G}_{\mathbb R} (n,k)$ and $\phi_E$ is a homeomorphism onto it's image. 
Consider the vector space $\mathbb R^n$ endowed with a positive definite inner product. Let $\mathcal {G}_{\mathbb R} (n,k)$ denote the set of all $k$-dimensional subspaces of $\mathbb R^n.$ Fix $E \in \mathcal {G}_{\mathbb R} (n,k)$ and define $U_{E} : = \left \{F \in \mathcal {G}_{\mathbb R} (n,k)\ \big |\ F \cap E^{\perp} = (0) \right \}.$ Consider the map $\phi_{E} : U_{E} \longrightarrow \mathcal L \left (E, E^{\perp} \right )$ defined as follows $:$
$$\phi_{E} (F) = P_{E^{\perp}} \circ P_{EF}^{-1},\ \ F \in U_{E}$$ where $P_{E^{\perp}}$ is the projection onto $E^{\perp}$ and $P_{EF} = P_{E} \rvert_{F},$ the projection onto $E$ restricted to $F.$ Show that
$(1)$ $U_{E}$ is an open subset of $\mathcal {G}_{\mathbb R} (n,k).$
$(2)$ $\phi_{E}$ is a homeomorphism from $U_{E}$ onto $\phi_{E} \left (U_{E} \right ).$
$(3)$ $\phi_{E} \circ \phi_{F}^{-1} : \phi_{F} \left (U_{E} \cap U_{F} \right ) \longrightarrow \phi_{E} \left (U_{E} \cap U_{F} \right )$ is a smooth map whenever $U_{E} \cap U_{F} \neq \varnothing.$

The above question has been left as an exercise in my lecture notes on Differential Geometry. The above exercise actually shows that $\mathcal {G}_{\mathbb R} (n,k)$ is a smooth manifold of dimension $k (n - k)$ with atlas $\left \{\left (U_{E}, \phi_{E} \right ) \right \}_{E\ \in\ \mathcal {G}_{\mathbb R} (n,k)}$ which is known as Grassmann manifolds. First I don't understand what is the topology on $\mathcal {G}_{\mathbb R} (n,k).$ This is not explicitly mentioned in the notes. For the second part I tried by taking an  $\varepsilon$-ball around $P_{E^{\perp}} \circ P_{EF}^{-1}$ in the image of $\phi_{E}$ for some $F \in U_{E}.$ The pre-image of this ball is not quite easy to obtain. If $F_1$ lies in the pre-image of the ball then we have the following inequality involving operator norm $$\left \|P_{E^{\perp}} \circ \left (P_{EF}^{-1} - P_{EF_1}^{-1} \right ) \right \| \lt \varepsilon.$$ I don't have any idea how to proceed further.
Any help would be greatly appreciated.
 A: You cannot solve part $(1)$ and $(2)$ of the excercise without knowing what the definition of the topology on  $\mathcal {G}_{\mathbb R} (n,k)$ is. But there is a way arround, maybe it helps:
First show:
$\bullet$ The $\phi_{E} : U_{E} \longrightarrow \mathcal L \left (E, E^{\perp} \right )$ are bijections. $\textbf{Hint:}$ $\phi_{E}^{-1}(A)=\mathrm{im}(A+\mathrm{id}_{|E})$.
$\bullet$ The $\phi_{E} \left (U_{E} \cap U_{F} \right )$ are open. $\textbf{Hint:}$ $A\in \mathcal L \left (E, E^{\perp} \right )$ lies in $\phi_{E} \left (U_{E} \cap U_{F} \right )$ iff $\mathrm{im}(A+\mathrm{id}_{|E})\cap F^{\perp}=\{0\}$.
$\bullet$ The $\phi_{E} \circ \phi_{F}^{-1} : \phi_{F} \left (U_{E} \cap U_{F} \right ) \longrightarrow \phi_{E} \left (U_{E} \cap U_{F} \right )$ are smooth.
Then theses three points plus the fact that the $U_E$ cover  $\mathcal {G}_{\mathbb R} (n,k)$ imply that there is a unique topology on $\mathcal {G}_{\mathbb R} (n,k)$ such that all $\phi_{E}$ become homeomorphisms. It is given by declaring $U\subseteq \mathcal {G}_{\mathbb R} (n,k)$ to be open iff all $\phi_E(U\cap U_E)$ are open. You may also want to show that this topology is second-countable and hausdorff.
For an alternative definition of $\mathcal {G}_{\mathbb R} (n,k)$ without charts you could indentify a $k$-dimensional subspace of $\mathbb R^n$ with the ortogonal projection onto it. Then $\mathcal {G}_{\mathbb R} (n,k):=\{A\in \mathrm{GL}(n,\mathbb R)\,|\,A^2=A^*=A, \mathrm{tr}(A)=k\}$ is a smooth submanifold of $\mathrm{GL}(n,\mathbb R)$ and the identification with your prevoius definition is a diffeomorphism.
