# Are morphisms between projective spaces required to be injective?

From Wikipedia's morphisms between projective spaces:

Injective linear maps $T \in L(V,W)$ between two vector spaces $V$ and $W$ over the same field $k$ induce mappings of the corresponding projective spaces $P(V) \to P(W)$ via: $$[v] \to [T(v)],$$ where $v$ is a non-zero element of $V$ and [...] denotes the equivalence classes of a vector under the defining identification of the respective projective spaces. Since members of the equivalence class differ by a scalar factor, and linear maps preserve scalar factors, this induced map is well-defined. (If $T$ is not injective, it will have a null space larger than $\{0\}$; in this case the meaning of the class of $T(v)$ is problematic if $v$ is non-zero and in the null space. ...).

1. In "if $T$ is not injective, the meaning of the class of $T(v)$ is problematic if $v$ is non-zero and in the null space", I wonder what kind of problem that is?
2. Are morphisms between projective spaces, projective linear transformation, and projective transformation (homography) different names for the same concept?

Thanks and happy holliday!

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Suppose that $T$ is not injective. Then there exists a nonzero $v$ in $V$ with $T(v) =0$ in $W$. So it has no "class" in $P(W)$. The elements in $P(W)$ are equivalence classes of elements in $W-\{0\}$. Did I misunderstand the question and am I telling you something you already knew? – Ohdur Dec 24 '11 at 23:01
The problem is simply that in projective space there is no origin. If $T(v) = 0 \in W$, then what is the corresponding image of $[v]$ in $P(W)$? If $T$ is injective then $0$ is the only point mapping to $0$, you remove this point on both sides of the map, and everything is fine. If you are thinking of projective space as a manifold or something else then there are certainly non-injective maps: $\mathbf RP^1$, for example, is a circle, and that has a lot of non-injective self-maps. – Dylan Moreland Dec 24 '11 at 23:02
@Ohdur: No, you didn't. Thanks! I forgot that the equivalent classes are in W−{0} not W. Are morphisms between projective spaces, projective linear transformation, and projective transformation (homography) different names for the same concept? – Tim Dec 24 '11 at 23:03
@DylanMoreland: Thanks! – Tim Dec 24 '11 at 23:11
@Tim. Map $(x,y)$ to $(x+y,x+y)$. This corresponds to the $(2\times 2)$-matrix with 1's everywhere on the standardbasis. This map has a one-dimensional kernel. It is spanned by the vector $(1,-1)$. – Ohdur Dec 24 '11 at 23:13

The null vector does not represent a valid element of a projective space. If $T$ were not injective, then there would be some $v$ which itself is non-zero but for which $T(v)$ is zero. In that case, $[v]$ is an element of $P(V)$ but $[T(v)]$ is not an element of $P(W)$, thus breaking the definition of the morphism.
(1) If $T$ is not injective, then it simply induces a projective map from $P(V)\setminus P(\ker T)$, which is an open set in $P(V)$, to $P(W)$. The easiest examples are provided by the very projective maps, namely, the central projections from a projective space onto a smaller subspace.