We proved the following theorem : Assume that $\psi:M^c\longrightarrow N^d$ is $C^{\infty}$. Let $n\in N$, $P=\psi^{-1}(n)$ be non-empty, and let $d\psi:M_m\longrightarrow N_{\psi(m)}$ is surjective for all $m\in P$. Then $P$ has a unique manifold structure such that $(P,i)$ is a submanifold of $M$, where $i$ is the inclusion map. Moreover, $i:P\longrightarrow M$ is a actually an imbedding, and the dimension of $P$ is $c-d$.
As an application of this we have been asked to prove that the orthogonal group $O(d)$ has a unique manifold structure such that $(O(d),i)$ is a submanifold of $Gl(d,\mathbb{R})$ using the map $\psi:Gl(d,\mathbb{R})\longrightarrow S^d$, where $S^d$ is the vector space of all $d\times d$ symmetric real matrices, and the map $\psi$ is given by $\psi(A)=AA^t$.
I have tried the following. Clearly, by the map, $O(d)=\psi^{-1}(I)$, where $I$ is the $d\times d$ identity matrix. So in order to apply the theorem and conclude, it is enough to check that $d\psi_{\sigma}$ is surjective at each $\sigma\in O(d)$.
But if $\sigma\in O(d)$, then $\psi=\psi\circ\tau_{\sigma}$, where $\tau_{\sigma}$ is given by $\tau_{\sigma}(A)=A\sigma$ (because $\psi\circ\tau_{\sigma}(A)=\psi\circ(A\sigma)=\psi(A)\circ\psi(\sigma)=\psi(A))$. But, this means that $d\psi_{I}=d(\psi\circ\tau_{\sigma})_I=d\psi_{\tau_{\sigma}(I)}\circ d\tau_{\sigma_{I}}=d\psi_{\sigma}\circ d\tau_{\sigma_{I}}$. Here, $\tau_{\sigma}$ is a diffeomorphism of $O(d)$, hence $d\tau_{\sigma_I}$ is an isomorphism.
So in order to check that $d\psi_{\sigma}$ is surjective for each $\sigma\in O(d)$, it is enough to check that $d\psi_I$ is surjective. But I am unable to prove that $d\psi_I$ is surjective. How can I prove this? Any help will be appreciated.
Note that $d\psi_I : M_{d\times d}(\mathbb R) \to S^d$ is given by
$$d\psi_I (X) = \frac{d}{ds} e^{sX} (e^{sX})^t \bigg | _{s=0} = X + X^t$$
Obviously the vector space $S^d$ is spanned by
$$E_{ii}, E_{ij} + E_{ji}$$
where $E_{ij}$ is the matrix with only nonzero entry at $(i, j)$. So we have
$$d\phi _I (\frac 12 E_{ii}) = E_{ii},\ \ d\psi_I (E_{ij}) = E_{ij} + E_{ji}$$
Thus $d\psi_I$ is onto.