What is this kind of map called? I am dealing with a linear map of the form
$$T: \mathbb{R}^3 \to \mathbb{R}^2: (x,y,z) \mapsto (x,y).$$
My source calls it an orthogonal projector onto $\mathbb{R}^2$ but that is misleading. An orthogonal projector would need to satisfy $T^\dagger = T$ and $T^2 = T$, while
$$T^\dagger: \mathbb{R}^2 \to \mathbb{R}^3: (x,y) \mapsto (x,y,0)$$
is a different operator and the latter condition is ill-defined for $T$.
How to properly call $T$ (and/or $T^\dagger$)?
 A: Strictly speaking you're right. However, we can view $\Bbb R^2$ as a subset of $\Bbb R^3$ by identifying ${\bf x} := (x, y) \in \Bbb R$ with its image under the injective map $T^{\dagger}({\bf x}) = (x, y, 0)$. Under this identification, we identify $T$ with the map $T^{\dagger} T : (x, y, z) \mapsto (x, y, 0)$, and we can check that this map is an orthogonal projection.
We can still, however, call $T : (x, y, z) \mapsto (x, y)$ a projection. If we canonically identify $\Bbb R^3$ with the Cartesian product $\Bbb R^2 \times \Bbb R$, i.e., via the map $(x, y, z) \mapsto ((x, y), z)$, $T$ is just the canonical projection onto the first factor. Likewise, again under this identification, $T^{\dagger}$ is the canonical inclusion of that factor. (While any Cartesian product admits canonical projections, the fact that this inclusion is canonical requires a little more: Namely, it requires a distinguished point in $\Bbb R$. Since we're working with vector spaces, this role is filled by the origin, $0$.)
