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
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 $\mathbb R^k$.