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inclusion mapping
Definition Let $X$ be a subset of $Y$. Then the inclusion map from $X$ to $Y$ is the mapping
$\displaystyle\iota:X$  $\displaystyle\to$  $\displaystyle Y$  
$\displaystyle x$  $\displaystyle\mapsto$  $\displaystyle x.$ 
In other words, the inclusion map is simply a fancy way to say that every element in $X$ is also an element in $Y$.
To indicate that a mapping is an inclusion mapping, one usually writes $\hookrightarrow$ instead of $\to$ when defining or mentioning an inclusion map. This hooked arrow symbol $\hookrightarrow$ can be seen as combination of the symbols $\subset$ and $\to$. In the above definition, we have not used this convention. However, examples of this convention would be:

Let $\iota:X\hookrightarrow Y$ be the inclusion map from $X$ to $Y$.

We have the inclusion $S^{n}\hookrightarrow\mathbb{R}^{{n+1}}$.
Related:
Pullback2
Synonym:
inclusion map, inclusion
Major Section:
Reference
Type of Math Object:
Definition
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Comments
Why is inclusion map a topological embedding?
I read in John M. Lee's book: Introduction to topological manifolds, proposition 3.4 (a) says that inclusion map is a topological embedding. But let's suppose there is an inclusion map from a closed subset A of X to X, then in the subspace of A inherited from X, A itself is an open set, while its image after the inclusion map is a closed subset in X. Henceforthly, this inclusion map between A and its image is not a homeomorphism. Hence no topological embedding. Where have I gone wrong? Please! Thank you!
Re: Why is inclusion map a topological embedding?
I assume that you define "topological embedding" as a map that yields a homeomorphism between its domain and its image (where the image carries the subspace topology). Then you're wrong and Mr. Lee is right: every inclusion map is a topological embedding.
In your example, if A is not open in X, then indeed A is open in the subspace topology of A, but A is also open in the (subspace topology) of the image of the inclusion, so there's no problem.
Your example has another flaw: you seem to think that "closed" is the opposite of "open" and that closed sets can never be open. That's not true.
If with "topological embedding" you mean an injective continuous and open map, then you are correct: not all inclusion maps are topological embeddings in this sense. The inclusion map of A into X is a topological embedding in this sense if and only if A is open.
Re: Why is inclusion map a topological embedding?
Thanks so greatly! I was troubled the whole night! Thank you!
inclusion map on Riemannian manifolds
My professor is pretending that the inclusion map i:M C Rn takes the points of definition (x1,...,xm) of a parametrization x:U > M (where M C Rn is a manifold of dimension m and U an open set of Rm) to Rn in the following sense: if m<n and p=x(q) is a point of M then i(p)=(x1,...,xm,0,...,0) is a point of Rn. This definition gives no sense to me. But I dont dare asking him again about this. Can anyone explain the sense of this definition and how it is compatible with the definition on this site?
Re: inclusion map on Riemannian manifolds
It seems to me he's just used the Rank Theorem for manifolds (which I state for your convenience) to give a canonical "flattened out" form to the inclusion map:
Given (1) smooth manifolds M, N of dimensions m, n resp.
(2) a smooth map f : M > N of constant rank k
(3) any point p in M
Then, there exist charts (G, g) and (H, h) "centred on" p and f(p) resp., f(G) contained in H such that the "coordinate representative" F of f w.r.t. the charts, acts as
(x_1, ..., x_k, ..., x_m) > (x_1, ..., x_k, 0, ..., 0).
[i.e. F : g^(1)(G) \subset IR^m > h(H) \subset IR^n, and F = hfg^(1).]
In your case, f = i, k = m, while U = g^(1)(G) and H have been quietly chosen as per the theorem. U is thus not just any parameter domain.
Re: inclusion map on Riemannian manifolds
That's an interesting theorem. Maybe my prof was thinking to that. By the why you made an error in your notation: if F = hfg^(1) then
F : g(G) \subset IR^m > h(H) \subset IR^n.
It is also possible that my prof was confusing the two definitions of a submanifold:
1. If the inclusion i:M>N is an embedding then M is a submanifold.
2. A subset M of a manifold N with dimension n is a submanifold of dimension m if there is a chart h:N>Rn with a coordinate neighbourhood H in N such that
h(H intersection M) = {x of h(H)  x(m+1)=x(m+2)=...xn=0}.
I think my prof is confusing i and h, which are not the same. I guess to prove the equivalence of these two definitions is a bit more complicate than just to equal i and h. I am wondering how this can be done. Any suggestions?
Re: inclusion map on Riemannian manifolds
In your 2nd definition, it should be a collection of "slice" charts covering M, because if the submanifold is compact, then of course, no single chart (of N) can cover N.
The definition of (smooth) embedding I use is : injective immersion + topological embedding.
Defn. (1) => Defn. (2) :
Let (G, g) be a generic slice chart for M. An atlas for N is given by the collection {pg : G intersect N > IR^m}, where p : IR^n > IR^m is the canonical projection. With this smooth structure, it is straightforward to show that i : N > M is a topological embedding and an injective immersion (in fact, it can also be shown that the smooth structure so obtained is the unique one making i a smooth embedding).
Defn. (2) => Defn. (1) :
Since i : N > M is a a smooth embedding, the rank theorem can be used to obtain charts u : U subset IR^m > V subset IR^n of the form (x_1, ..., x_m) > (x_1, ..., x_m, 0, ..., 0). These charts can, after possibly some shrinking of domain, be used to easily obtain slice charts for N.
Details : see John M. Lee, Introduction to Smooth Manifolds (Springer 2003).