The differential of the inclusion map is the inclusion map of tangent spaces. Let $X$ and $Y$ be smooth manifolds, let $i:X\to Y$ be the inclusion map, prove $di_x$ is the inclusion map from $T_x(X)$ to $T_x(Y)$. I know this is pretty basic, but can someone show me how to do this? The definition of derivatives on manifold are messing with me.
Thanks 
 A: Let me provide some details regarding what I think the setup of your problem is, and hopefully that will clarify some things for you. If $i : X \to Y$ is an inclusion map, that implies that $X$ is a submanifold of $Y$. So suppose dim $Y = n$ and dim $X = k \leq n$. There are two ways to view $X,$ either 1) as a subset of $Y$ or 2) as an abstract $k$ dimensional manifold. These two viewpoints are related via the inclusion map $i : X \to X \subset Y$. Since $X$ is a submanifold of $Y$, around any point $x \in X$ we may choose a diffeomorphism $\phi :U \to \mathbb R^n$, where $U \ni x$ is an open set in $Y$, with the property that the first $k$ factors of $\phi$ map $\mathbb R^k$ onto $U \cap X$, i.e. we may arrange $\phi$ so that $$(x_1, \ldots, x_k, 0,\ldots, 0)$$ is the coordinate representation of $U \cap X$, where $x_i$  ($1\leq i \leq k$) are the first $k$ coordinate functions of $\phi$. In particular the restriction of $\phi$ to the first $k$ factors provides a coordinate chart for $X$ as an abstract manifold. Thus the coordinate representation of $i$ is nothing but the map $$(x_1, \ldots, x_k) \mapsto 
(x_1, \ldots, x_k, 0, \ldots, 0).$$ Now take the derivative of this map at any point $x_0$ 
in the coordinate chart and note that at the point $x_0$ you may identify the tangent space of $X$ with $\mathbb R^k$.
