Embeddings in topology

In topology, an embedding is a continuous and injective function $ f: X \rightarrow Y $ between two topological spaces $ X $ and $ Y $, such that $ f $ induces a homeomorphism between $ X $ and its image $ f(X) $, where $ f(X) $ is given the subspace topology from $ Y $.

This means an embedding has three key properties:

The function $ f $ is continuous.
The function $ f $ is injective, meaning no two distinct points in $ X $ are mapped to the same point in $ Y $.
The inverse of $ f $, seen as a function from $ f(X) $ to $ X $, is continuous with respect to the subspace topology on $ f(X) $

In other words, an embedding preserves the topological structure of $ X $ within its image $ f(X) $ in $ Y $, allowing $ f(X) $ to be viewed as a subspace of $ Y $, i.e., $ f(X) \subset Y $.

A practical example
The difference between an embedding and a homeomorphism

A practical example

Let’s define two topological spaces:

Space $ X $
The set $ X = \{a, b, c\} $ with the topology $ \mathcal{T}_X = \{\emptyset, \{a\}, \{a, b\}, X\} $, which defines the open sets in $ X $.
Space $ Y $
The set $ Y = \{1, 2, 3, 4\} $ with the topology $ \mathcal{T}_Y = \{\emptyset, \{1\}, \{1, 2\}, \{1, 2, 3\}, Y\} $, which defines the open sets in $ Y $.

Now, define a function $ f: X \rightarrow Y $ as follows:

$$ f(a) = 1 \\ f(b) = 2 \\ f(c) = 3 $$

Next, let’s check if $ f $ satisfies the conditions for an embedding.

1] Continuity of $ f $

A function $ f: X \rightarrow Y $ is continuous (see the definition of continuity with open sets) if the preimage of every open set in $ Y $ is an open set in $ X $, meaning the preimage belongs to the topology $ \mathcal{T}_X $.

$ f^{-1}(\emptyset) = \emptyset \in \mathcal{T}_X $, which is open in $ X $.
$ f^{-1}(\{1\}) = \{a\} \in \mathcal{T}_X $, which is open in $ X $.
$ f^{-1}(\{1, 2\}) = \{a, b\} \in \mathcal{T}_X $, which is open in $ X $.
$ f^{-1}(\{1, 2, 3\}) = X \in \mathcal{T}_X $, which is open in $ X $.
$ f^{-1}(Y) = X \in \mathcal{T}_X $, which is open in $ X $.

Since the preimage of every open set in $ \mathcal{T}_Y $ is open in $ \mathcal{T}_X $, the function $ f $ is continuous.

2] Injectivity

The function $ f $ is injective because it maps each element of $ X $ to a distinct element in $ Y $.

$$ f(a) = 1 \\ f(b) = 2 \\ f(c) = 3 $$

3] Continuity of the inverse

Next, consider the image of $ f $, which is $ f(X) = \{1, 2, 3\} \subset Y $.

The subspace topology on $ f(X) $ is $ \mathcal{T}_{f(X)} = \{\emptyset, \{1\}, \{1, 2\}, \{1, 2, 3\}\} $.

Note. The subspace topology on a subset of a topological space is defined as the collection of all sets that are intersections of open sets from the original space with the subset. In this case:

The space $ Y = \{1, 2, 3, 4\} $ has the topology $ \mathcal{T}_Y = \{\emptyset, \{1\}, \{1, 2\}, \{1, 2, 3\}, \{1, 2, 3, 4\}\} $.
The image of $ f(X) $ is $ f(X) = \{1, 2, 3\} \subset Y $.

The possible intersections are:

$ \emptyset \cap \{1, 2, 3\} = \emptyset $
$ \{1\} \cap \{1, 2, 3\} = \{1\} $
$ \{1, 2\} \cap \{1, 2, 3\} = \{1, 2\} $
$ \{1, 2, 3\} \cap \{1, 2, 3\} = \{1, 2, 3\} $
$ \{1, 2, 3, 4\} \cap \{1, 2, 3\} = \{1, 2, 3\} $ (which coincides with the previous set)

Therefore, the subspace topology $ \mathcal{T}_{f(X)} $ is: $$ \mathcal{T}_{f(X)} = \{\emptyset, \{1\}, \{1, 2\}, \{1, 2, 3\} \} $$

To check the continuity of the inverse restricted to the image $ f^{-1}: f(X) \rightarrow X $, we need to verify that the preimage of each open set in $ \mathcal{T}_X $ is open in the subspace topology $ \mathcal{T}_{f(X)} $.

$ \emptyset $ is open in $ \mathcal{T}_X $, and its preimage under $ f^{-1} $ is $ \emptyset $, which is open in $ \mathcal{T}_{f(X)} $.
$ \{a\} $ is open in $ \mathcal{T}_X $, and its preimage under $ f^{-1} $ is $ \{1\} $, which is open in $ \mathcal{T}_{f(X)} $.
$ \{a, b\} $ is open in $ \mathcal{T}_X $, and its preimage under $ f^{-1} $ is $ \{1, 2\} $, which is open in $ \mathcal{T}_{f(X)} $.
$ X $ is open in $ \mathcal{T}_X $, and its preimage under $ f^{-1} $ is $ \{1, 2, 3\} $, which is open in $ \mathcal{T}_{f(X)} $.

Thus, the inverse function $ f^{-1} $ restricted to the image $ f(X) $ is continuous.

In conclusion, the function $ f $ is an embedding between $ X $ and $ f(X) $ because it is continuous, injective, and its inverse, restricted to the image $ f(X) $, is continuous.

Even though the image of $ X $ under $ f $ ($ \{1, 2, 3\} $) does not cover all of $ Y $, it preserves the topological structure of $ X $ within its image.

The difference between an embedding and a homeomorphism

The difference between an embedding and a homeomorphism lies in the domain, the image of the function, and how they preserve the topological structure.

Homeomorphism
A homeomorphism is a bijective function that maps $ X $ onto $ Y $, preserving the topological structure of both spaces entirely.
Embedding
An embedding is a function that maps $ X $ into $ Y $ in such a way that the topological structure of $ X $ is preserved, but only within the image $ f(X) $, which is a subspace of $ Y $.

In other words, while a homeomorphism implies a full one-to-one correspondence between two spaces, an embedding only ensures that $ X $ corresponds to a subspace of $ Y $, with the structure of $ X $ preserved within that subspace $ X \subset Y $.

And so on...