exp is a homeomorphism from symmetric to positive definite matrices

Idea. $\exp$ is a continuous bijection; we show it is also a closed map by checking properness.

Well defined. For $S\in\mathrm{Sym}_n$ there is $Q\in O(n)$ with $S=Q\,\mathrm{diag}(\lambda_i)\,Q^T$, so $e^S=Q\,\mathrm{diag}(e^{\lambda_i})\,Q^T$, which is symmetric with eigenvalues $e^{\lambda_i}>0$: positive definite.

Bijective. Define $\log P=Q\,\mathrm{diag}(\ln\mu_i)\,Q^T$ for $P=Q\,\mathrm{diag}(\mu_i)\,Q^T\in\mathrm{PD}$. This does not depend on the choice of diagonalisation: if $f$ is any function and $P$ has eigenspaces $E_\mu$, then $f(P)$ is the operator acting as the scalar $f(\mu)$ on each $E_\mu$, and this description is intrinsic. By construction $\exp(\log P)=P$ and $\log(\exp S)=S$. So $\exp$ is a bijection $\mathrm{Sym}_n\to\mathrm{PD}$ with inverse $\log$.

Continuous. $\exp$ is continuous on $M_n(\mathbb{R})$ and $\mathrm{Sym}_n$ carries the induced topology.

Proper. Let $K\subset\mathrm{PD}$ be compact. Eigenvalues depend continuously on the matrix (e.g. $\mu_{\max}(P)=\max_{\|x\|=1}x^TPx$ and $\mu_{\min}(P)=\min_{\|x\|=1}x^TPx$ are continuous), so there are $0<a\le b<\infty$ with all eigenvalues of every $P\in K$ in $[a,b]$. Then for $S=\log P$ the eigenvalues $\ln\mu_i$ lie in $[\ln a,\ln b]$, so $\|\log P\|_{\mathrm{op}}\le\max(|\ln a|,|\ln b|)$ for $P\in K$: the set $\exp^{-1}(K)\subset\mathrm{Sym}_n$ is bounded. It is closed: $\exp^{-1}(K)=\{S\in\mathrm{Sym}_n:e^S\in K\}$ is the preimage of the compact, hence closed, subset $K\subset M_n(\mathbb{R})$ under the continuous map $\exp$. A closed bounded subset of $\mathrm{Sym}_n\cong\mathbb{R}^{n(n+1)/2}$ is compact.

Conclusion. The inverse $\log=\exp^{-1}$ is continuous: if $F\subset\mathrm{Sym}_n$ is closed, then $\log^{-1}(F)=\exp(F)$. To see that $\exp(F)$ is closed in $\mathrm{PD}$, take $P_k=e^{S_k}\to P\in\mathrm{PD}$ with $S_k\in F$. The set $\{P_k\}\cup\{P\}$ is compact in $\mathrm{PD}$, so by properness $(S_k)$ lies in a compact set; a subsequence converges to some $S\in F$ ($F$ closed), and $e^S=\lim e^{S_{k_j}}=P$. So $P\in\exp(F)$.

Remark. The same proof shows that $\exp$ is a homeomorphism between the open cone $\mathrm{PD}$ (convex, hence contractible) and the vector space $\mathrm{Sym}_n$. It is even a diffeomorphism (it is smooth with smooth inverse, by the holomorphic functional calculus).