How to show that $f:[0,1]\times{\rm Sp}(n,\mathbb{R})\to {\rm Sp}(n,\mathbb{R})$ is continuous? If $f:[0,1]\times {\rm Sp}(n,\mathbb{R})\to {\rm Sp}(n,\mathbb{R})$ is defined as
$$f(t,A)=(A^{T}A)^{-t/2}$$
where, ${\rm Sp}(n,\mathbb{R})$ denotes the set of $n\times n$ real symplectic matrices. Prove that $f$ is continuous.
As far my knowledge, I know only three methods to prove a function is continuous:
(i) A function $f:X\to Y$ between topological spaces $X$ and $Y$ is continuous if and only if for every open(closed) subset $A\subset Y$, $f^{-1}(A)$ is open(Closed),
(ii) $\epsilon-\delta$ definition and
(iii) Composition of continuous function is continuous.
I tried using any of these methods but I cannot, for (i) I dont know what type of sets in ${\rm Sp}(n,\mathbb{R})$ are closed or open, for (ii) I dont know what the metric is and (iii) I'm still figuring out..
 A: Note that $A^TA$ being a positive definite matrix , its logarithm is well defined, as the unique (real) matrix $B$ such that $e^B = A^TA$. You can verify this from Wikipedia, 
The function $B \to \log B$ is continuous on symmetric positive definite matrices. This can be explained by the following logic : $A$ being positive definite and symmetric, one easily sees by the spectral theorem that  $B = UDU^{-1}$ (for $D$ diagonal consisting of the eigenvalues of $B$ and $U$ the matrix of eigenvectors of $B$), we have $\log B = U (\log D) U^{-1}$ where $U$ is the matrix consisting of eigenvectors of $B$ and $\log D$ is taking the logarithm of every diagonal entry of $D$.
Once this happens, note that as $B_n \to B$ in matrix norm in this space of psd matrices, the eigenvalues converge, and therefore so do the eigenvectors (only in this case, not in general, because if $B_n = U_nD_nU_n^{-1}$, $B_n$ is not too far from $B$, $D_n$ is not too far from $D$ so $U_n$ can't be too far from $U$ etc. Read more on this kind of estimation here.) in any norm of choice. It follows by continuity of the real logarithm, that the matrix logarithm is continuous.
That, composed with $A \to A^TA$ being continuous, tells you that $A \to \log A^TA$ is a continuous function.
Finally, $t\to -\frac t2$ is continuous obviously. Now, a product of continuous functions of separate variables is continuous in the joint topology, so $(t,A) \to (-\frac t2 \log(A^TA))$ is continuous. 
The exponential is continuous, being a uniform limit of the polynomials $P_n(T) = \sum_{k=0}^n \frac{T^k}{k!}$. Thus, after composition , we find that $$(t,A) \to \exp\left(-\frac t2 \log(A^TA)\right)$$
is continuous.
The RHS equals $(A^TA)^{-\frac t2}$ by definition of a power of a matrix when the power is not rational. Thus, by a series of compositions (with some black boxes for you to fill) the result is a continuous function.
