Showing the inclusion map of a submanifold is an immersion. In the recent course of studying manifolds by the lecture note provided by Ed Segal of the UCL(http://www.homepages.ucl.ac.uk/~ucaheps/papers/Manifolds%202016.pdf), I encountered a question asking me to show that, for a submanifold Z of X the inclusion map $i: Z\hookrightarrow X$ is an immersion.
And I don't know how to show this. I first tried by envisaging a coordinate chart to the standard Euclidean space such that $Di|_x=(I|0)$, but I am not sure this is the right approach.
Could somebody please help me solve this problem?
c.f. The author of the lecture note defines the "submanifold" of a topological space X as a subset of X whose open sets are mapped by the chart function $f$ to the subspace topology of a certain affine subspace A.

Let $Z$ be a k-dimensional submanifold of $X$. Assume that the dimension of $X$ is $n\geq k$. Now, let $p\in Z$. As $Z$ is a submanifold, there exist a chart $(U_p,\phi)$ around $p$ with $\phi:U_p \rightarrow \mathbb{R}^k\times\mathbb{R}^{n-k}$, such that
$$\phi\big|_{Z}:U_p\cap Z\rightarrow \mathbb{R}^{k}\times\{0\}.$$
is a diffeomorphism onto its image. Hence $(U_p\cap Z,\phi\big|_{Z} )$ gives a local chart for $Z$ around $p$. With this local chart, we can compute $Di_p$ for the inclusion map $i:Z\rightarrow X$. Take $(U_p\cap Z,\phi\big|_{Z} )$ as a chart around $p$ in $Z$ and $(U_p,\phi)$ as a chart around $p$ in $X$. Then the rank of $i$ (by which I mean the rank of $Di_p$) is the same as the rank of the map,
$$j:\mathbb{R}^k\rightarrow \mathbb{R}^k\times \mathbb{R}^{n-k}$$
$$j(x_1,...,x_k)= (x_1,...,x_k,0,..,0)$$
as
$$i =\phi^{-1}\circ j\circ \phi\big|_{Z}$$
The map $j$ obviously has rank $k$ at every point. Hence $i$ has rank $k$ at $p$.
