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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

2 Answers 2

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.

Ohdur wrote most of this already in his comment, so there is nothing new in this answer, except the fact that it is technically an answer and not just a comment.

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(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.

(2) The terminology is quite wild, however, the two names you mention do refer to the same thing.

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