A flat torus showing simple connectedness is needed for global exp injectivity

The quotient map $q$ is a local isometry, a covering map, and $(T^n,g)$ is complete. [F1, F3, F8, given] In a periodic chart of [F1] the metric matrix is the identity. The local inverse charts of $q$ are the inverses of the restrictions of $q$ to boxes of side lengths less than $1$, and in those coordinates $q$ is the identity map of an open subset of ${\mathbb{R}}^n$; hence the differential of $q$ preserves the Euclidean metric of ${\mathbb{R}}^n$ and the flat metric of $T^n$ at every point. By [F3] this makes $q$ a local isometry from the boundaryless connected Riemannian manifold $({\mathbb{R}}^n,g_{\mathrm{E}})$ onto $(T^n,g)$, where $T^n$ is connected and boundaryless by [F1]. The Euclidean space is complete by [F8] and nonempty since $n\ge 2$. The local isometry is therefore a covering map by [F3], and part 3 of [F3] makes $(T^n,g)$ complete.

The torus has constant sectional curvature $K=0$. [F2, step 1.1] Take a periodic chart of [F1] with coordinates $x^1,\dots ,x^n$, in which the metric matrix is constantly $I_n$. Its first derivatives vanish identically, so every Christoffel symbol vanishes by the formula of [F2]. Substituting the vanishing symbols into the coordinate formula of [F2] gives $R^{\mathrm{\ell }}{}_{kij}=0$ in the chart, that is, the Riemann curvature four-tensor vanishes there, and since the charts cover $T^n$ it vanishes on all of $T^n$. Hence $\mathrm{Rm}(X,Y,Y,X)=0$ for every tangent pair, and dividing by the Gram determinant, which is positive for an independent pair by [F2], gives $K=0$ at every tangent two-plane; in particular $K\le 0$.

Straight lines are the geodesics and the exponential map is $v\mapsto [x+v]$. [F2, F4, step 1.1] Fix $x,v\in {\mathbb{R}}^n$ and let $\gamma (t)=q(x+tv)$ for $t\in \mathbb{R}$. In a periodic chart containing $q(x+tv)$ the coordinate expression of $\gamma$ is $t\mapsto x+tv$ up to a constant integer translation, and its coordinate acceleration is identically zero; since the Christoffel symbols of [F1] vanish in these charts, [F2] makes $\gamma$ a geodesic. Its initial point is $[x]$ and its initial tangent is the vector identified with $v$, and it is defined on all of $\mathbb{R}$; by uniqueness of the maximal geodesic with given initial data the world line is complete. Thus every fibrewise exponential map has domain all of $T_{[x]}{T}^n$ and ${\exp }_{[x]}(v)=\gamma (1)=[x+v]$, the formula recorded in [F4].

The exponential map is the quotient map and is not injective. [step 2.2] The chart identification of $T_{[x]}{T}^n$ with ${\mathbb{R}}^n$ is linear, so step 2.2 says that ${\exp }_{[x]}$ is exactly the quotient projection $q$ composed with the translation $v\mapsto x+v$; explicitly, $q(w)=[w]$ and ${\exp }_{[x]}(v)=q(x+v)$. Thus it is the quotient projection after translation, and is surjective. Let $z=e_1\in {\mathbb{Z}}^n$ be the first standard basis vector; this is a nonzero lattice vector because $n\ge 2$. The tangent vectors $0$ and $z$ are distinct, but step 2.2 gives ${\exp }_{[x]}(0)=[x]=[x+z]={\exp }_{[x]}(z)\mathrm{.}$ Hence ${\exp }_p$ is not injective for any $p=[x]\in T^n$.

The torus is not simply connected. [F5, F6, step 1.1, step 2.2] Define $\gamma :[0,1]\to T^n$ by $\gamma (s)=[s{e}_1]$, a based loop at $[0]$. Its lift starting at $0$ is $s\mapsto s{e}_1$, so its endpoint is $e_1\mathrm{\ne }0$ (Existence and uniqueness of path lifts through a covering map in [F5]). Suppose it were homotopic relative endpoints to the constant loop, via $H:[0,1{]}^2\to T^n$ with $H(s,0)=\gamma (s)$, $H(s,1)=[0]$, and $H(0,t)=H(1,t)=[0]$. Lift $H$ with $\tilde{H}(0,0)=0$ by [F5]. The lower edge lifts $\gamma$, so $\tilde{H}(1,0)=e_1$. The right edge lifts the constant path starting at $e_1$, hence is constantly $e_1$. The top edge lifts the constant path starting at $\tilde{H}(0,1)=0$, hence is constantly $0$. Thus $\tilde{H}(1,1)=e_1=0$, a contradiction. Therefore $[\gamma ]$ is not the identity in ${\pi }_1(T^n,[0])$. Since $T^n$ is nonempty and path connected, it is not simply connected by [F6].

Conclusion, and the hypotheses of Cartan–Hadamard. [F3, F7, step 2.1, step 3.1, step 3.2] By step 1.1 the torus is complete, by step 2.1 it has $K=0\le 0$, and it is a connected boundaryless Riemannian manifold; by step 3.2 it is not simply connected. It therefore satisfies every hypothesis of Cartan hadamard except simple connectedness, and by step 3.1 the conclusion of that theorem — injectivity of ${\exp }_p$ — fails at every point. Hence simple connectedness cannot be omitted from Cartan–Hadamard. The failure is exactly the deck-group phenomenon: for every $p$ the fibre of ${\exp }_p$ over $p$ is the lattice ${\mathbb{Z}}^n$ of step 3.1 translations, and the covering $q$ of step 1.1 is nontrivial precisely because the deck translations are nontrivial. Boundary cases: the word "constant curvature $0$" is two-plane curvature, so it is asserted only for $n\ge 2$ as in the statement; the vector $z=e_1$ used in step 3.1 is nonzero exactly because $n\ge 1$; the case $v=0$ of step 2.2 is the constant geodesic through $[x]$, and $0$ and $z$ are distinct vectors with the same image, so the failure of injectivity is not an artefact of a degenerate vector. No choice beyond the inherited [A1] is used: the lattice, the charts, the loop and the homotopy argument are all explicit.