Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to show that for X being an n-dimensional manifold and Y a k-dimensional manifold, U an open set and both $Y,U \subset X, p\in U$ there is a natural and well defined and injective map $$i_{*}^{alg}:T_{p}^{alg}Y \rightarrow T_{p}^{alg}X \text{ is defined by } i_{*}^{alg}(v)([f,U]) \equiv v([f|_{U \cap Y}, U \cap Y])$$

Now I am guessing that wehat I am needing to show is that $f|_{U \cap Y}: U \cap Y \rightarrow \mathbb{R}$ is differentiable, that $(f|_{U \cap Y}, U \cap Y) \text{ defines the germ } [f|_{U \cap Y}, U \cap Y]$ of a differentiable function on $Y \text{ near } p, \text{where } f:U \rightarrow \mathbb{R}$ is a differentiable function. But this is just the first step. Could I say that this is true because $g:Y \rightarrow \mathbb{R}^k = \mathbb{R}^{k-1} \times \mathbb{R}$ and so $U \cap Y$ is essentially $f \cap g: U \cap Y \rightarrow \mathbb{R} \cap (\mathbb{R}^{k-1} \times \mathbb{R}) = \mathbb{R}?$

Now to show the definition of $i_{*}^{alg}$ works, we would have to show that $i_{*}^{alg} \circ (v \circ([f,U]))$ is as defined above, but that $(v \circ([f,U])) \in T_{p}^{alg}Y$. Is that the key? Any thoughts on how I could show this?

Thanks in advance,

Brian

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.