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Let $X$ and $Y$ be Euclidean spaces and consider the tangent map of a smooth map $f:X\rightarrow Y$, $T_p f: T_pX \rightarrow T_{f(p)}Y$, defined by $(p, v) \mapsto (f(p), \partial f(p)v)$. The author of the text I am reading claims that the image of the tangent map is a subspace of $T_{f(p)}Y$ for the following reason:

Because $v \mapsto f(p) + \partial f(p)v$ approximates (up to first order) the map $f$ at the point $p$ when $v$ is in a null neighborhood of $\mathbb{R^n}$ we know that $Im(T_pf)$ is a vector subspace of $T_{f(p)}Y$, which approximates (to first order) $f(X)$ at the point $f(P)$

If fail to understand why it is necessary to invoke a Taylor expansion to show that the image is a subspace and, moreover, how this argument actually proves that it is a subspace in any event. I'm sure I'm missing something, but showing that $Im(T_pf)$ is a subspace seems to me to be just an easy exercise in applying the linearity of the derivative of $f$ at $p$. E.G., for two points in $T_{f(p)}Y$ we have

$$ (f(p), \partial f(p) u) + (f(p), \partial f(p) v) = (f(p), \partial f(p)u + \partial f(p)v) = T_pf(u+v) \in Im(T_p f) $$

Homogeneity is just as easy. So my questions are one, What point is the author trying to make by the provided argument and two, If my argument is not correct where does it fail?

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What do you mean by " a null neighborhood of $R^n$"? Is it just a translate of a neighborhood of the origin? –  gary Jun 16 '11 at 18:58
@gary A "null neighborhood" is a neighborhood of 0 –  ItsNotObvious Jun 16 '11 at 20:10
I don't think the author wants to prove that the image of the tangent map in $p$ is a subspace. I guess that the key phrase in what you've quote there is that the image of the tangent maps APPROXIMATES $f(X)$ around the point $f(P)$. The tangent map is linear, the tangent space is a linear space, and therefore the image of a linear space is a linear space. –  Beni Bogosel Jun 16 '11 at 21:34
@Beni You may be right but, if so, it's a terribly unclear statement. –  ItsNotObvious Jun 16 '11 at 22:18

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