Aut$(V)$ open subset of End$(V)$ I have read in a book that Aut$(V)$ (set of all invertible linear maps) is an open subset of the End$(V)$ (set of all linear maps), where $V$ is a vector space. Can someone explain under what topology(ies) it is the case?
 A: Let $V$ be a finite-dimensional vector space (the finite-dimensionality here is of utmost importance). By fixing a basis, you obtain a linear isomorphism $V\rightarrow\mathbb{R}^n$. Now we can take the Euclidean topology and transport is along this isomorphism to obtain a topology on $V$, which makes this isomorphism into a homeomorphism. This topology on $V$ does not depend on the choice of linear isomorphism $V\rightarrow\mathbb{R}^n$, because any other such isomorphism is obtained by composing the first one with a linear automorphism $\mathbb{R}^n\rightarrow\mathbb{R}^n$ and these are all homeomorphisms (check this). Thus we obtain a well-defined topology on $V$. One nice property of this topology is that it makes the addition $V\times V\rightarrow V$ and the scalar multiplication $\mathbb{R}\times V\rightarrow V$ (equip the product spaces with the product topologies) into continuous maps (check this as well). This seems like a desirable property and, in fact, it is possible to show that this is the unique Hausdorff topology on $V$ with this property, though that takes some more work. All in all, we obtain a canonical topology on $V$. It should be remarked that this is completely false in case $V$ is infinite-dimensional.
Anyway, if $V$ is a finite-dimensional vector space, so is $\mathrm{End}(V)$. In fact, we can choose a basis to identify $V$ with $\mathbb{R}^n$ as above and that naturally yields an identification of $\mathrm{End}(V)$ with the space $\mathbb{R}^{n\times n}$ of $n\times n$-matrices (both as vector spaces and as topological spaces, by the above construction). Now $\mathrm{Aut}(V)$ corresponds to the space $\mathrm{GL}(n)$ of invertible $n\times n$-matrices (as topological space, it carries the subspace topology inherited from $\mathbb{R}^{n\times n}$). From linear algebra, we know a matrix is invertible iff its determinant is non-zero. The determinant can be thought of as function $\det\colon\mathbb{R}^{n\times n}\rightarrow\mathbb{R}$ and so $\mathrm{GL}(n)=\det^{-1}(\mathbb{R}\setminus\{0\})$. Since $\mathbb{R}\setminus\{0\}\subseteq\mathbb{R}$ is open, $\mathrm{GL}(n)$ will be open as long as $\det$ is continuous. But that is clear from the Leibniz formula, which shows that $\det$ is in fact a polynomial.
A: For an infinite dimensional example, we have that the set of invertible bounded linear operators on a Banach space is open in the norm topology. This can be shown using Neumann series.
