Computing the dimension of $O(n)$ - why is this matrix invertible? I am looking at Lee's Smooth Manifolds second edition and on page 167 he computes the dimension of $O(n)$. Now there he defines a map 
$$\begin{eqnarray*}\Phi : &\text{GL}(n,\Bbb{R})& \to M(n,\Bbb{R})  \\
&A& \mapsto A^T A \end{eqnarray*}$$
and wants to compute the rank of $\Phi$. Now he says for any $B \in T_I\text{GL}(n,\Bbb{R})$ let $\gamma : (-\epsilon,\epsilon) \to \text{GL}(n,\Bbb{R})$ be the smooth curve $\gamma(t) = I + tB$ and so on.


My question is: Why should the matrix $I + tB$ be an invertible matrix for all $t \in (-\epsilon,\epsilon)$? My guess is this: at $t = 0$ the determinant of $I + (0\cdot B)$ is $1$. Now consider the map $$f(t)\stackrel{\text{def}}{\equiv} \det \gamma(t) : \Bbb{R} \to \Bbb{R}.$$ Now $f(0) = 1$ and because $\Bbb{R} - \{0\}$ is open I can choose a $\varepsilon > 0 $ so that $B_\varepsilon(1) \subseteq \Bbb{R} - \{0\}$. Then by continuity there exists $\delta > 0$ so that $f(B_\delta(0)) \subseteq B_\varepsilon(1)$, that is to say I can choose an $\delta > 0$ so that $\gamma(t)$ is invertible for all $t \in (-\delta,\delta)$.



You are correct.
The curve $\gamma$ you defined is continuous from $\mathbb{R}$ to $M(n, \mathbb{R})$ and the determinant function is continuous from $M(n, \mathbb{R})$ to $\mathbb{R}$.
So, the map $t \mapsto \operatorname{det}(\gamma(t))$ is continuous on $\mathbb{R}$, taking the value $1$ at $t = 0$. Therefore there is a small interval around $0$ on which $\operatorname{det}(\gamma(t))$ is nonzero.
