The term injective comes from linear transformations. Recall that an injective linear map $T : V \rightarrow W$ is one that for every $w$ in $W$, there is at most one $v$ in $V$ such that $T(v)=w$. If T is injective then we can define a projection $P : W \rightarrow U$ such that $P$ is a linear map and $\mathrm{ker}(P) \cap \mathrm{im}(T)=\{0\}$, and $P(T)$ is both injective and surjective, thus it's invertible.
For the case of the GHZ state over 2 qubits, we have:
$$
|\mathrm{GHZ}\rangle = \frac{1}{\sqrt{2}}\sum_{\{s_1,s_2\}\in\{0,1\}} \mathrm{Tr} (A^{[s_1]} A^{[s_2]}) |s_1s_2\rangle
$$
We can write the open boundary condition tensor for this state as follows:
$$
\mathrm{GHZ}_\mathrm{OBC} = \frac{1}{\sqrt{2}}\sum_{\{s_1,s_2\}\in\{0,1\}} A^{[s_1]} A^{[s_2]}
$$
Which is a 4-tensor. We can write out this 4-tensor as (up to normalization):
$$
A_{{s_1}{s_2}}^{{a_1}{a_2}} = \delta_{s_1}^0\delta_{s_2}^0\delta_{a_1}^0\delta_{a_2}^0 + \delta_{s_1}^1\delta_{s_2}^1\delta_{a_1}^1\delta_{a_2}^1
$$
We can construct a projector that contracts two of the indices:
$$
(PA)_{s_2}^{a_2} = A_{{s_1}{s_2}}^{{s_1}{a_2}}
$$
Using Einstein notation. You can see that this projector results in the identity tensor, which is its own inverse.