Pasting lemma

Let $ X $ be a topological space with two closed subsets $ A $ and $ B $ such that their union covers the entire space, i.e., $ A \cup B = X $. If the functions $ f: A \to Y $ and $ g: B \to Y $ are continuous into a topological space $ Y $, and they agree on the overlap, meaning $ f(x) = g(x) $ for every point in the intersection $ x \in A \cap B $, then the function $ h: X \to Y $ defined by: $$ h(x) = \begin{cases} f(x) & \text{if } x \in A, \\ g(x) & \text{if } x \in B, \end{cases} $$ is continuous.

In simple terms, under certain conditions, two functions can be "glued" together to form a new continuous function.

Given two continuous functions $ f: A \to Y $ and $ g: B \to Y $, defined on overlapping sets $ A $ and $ B $ that agree on their intersection, we can "glue" them to form a new continuous function $ h $ defined on the union $ A \cup B $.

A practical example
Proof

A practical example

Let's look at two functions defined on different intervals:

$ f: [0, 1] \to \mathbb{R} $, defined as $ f(x) = x $, which is continuous on $ [0, 1] $,
$ g: [1, 2] \to \mathbb{R} $, defined as $ g(x) = 2 - x $, which is continuous on $ [1, 2] $.

Now, let’s check whether the conditions of the pasting lemma are met:

Closed sets: The intervals $ [0, 1] $ and $ [1, 2] $ are closed in $ \mathbb{R} $.
Covering the interval: The union of $ A = [0, 1] $ and $ B = [1, 2] $ is the interval $ [0, 2] $, so $ A \cup B = [0, 2] $.
Agreement at the overlap: The intersection is $ A \cap B = \{1\} $. Let's check that $ f(1) = g(1) $:
- $ f(1) = 1 $
- $ g(1) = 2 - 1 = 1 $
Therefore, $ f(1) = g(1) = 1 $, so the continuity condition at the point of overlap is satisfied.

All the conditions of the lemma are satisfied.

Now we define the function $ h: [0, 2] \to \mathbb{R} $ as:

$$ h(x) = \begin{cases} f(x) = x & \text{if } x \in [0, 1], \\ g(x) = 2 - x & \text{if } x \in [1, 2]. \end{cases} $$

The function $ h $ is continuous because:

On $ [0, 1] $, $ h(x) $ matches $ f(x) = x $, which is already known to be continuous on this interval.
On $ [1, 2] $, $ h(x) $ matches $ g(x) = 2 - x $, which is also continuous on this interval.
At the point of overlap $ x = 1 $, we’ve already verified that $ f(1) = g(1) = 1 $, so $ h(x) $ is continuous at $ x = 1 $ as well.

Since both $ f $ and $ g $ are continuous on their respective intervals and agree at the point of overlap, the function $ h(x) $ is continuous across the entire interval $ [0, 2] $.

Overall, the function $ h(x) $ is composed of two segments:

On the interval $ [0, 1] $, $ h(x) = x $, which is a straight, increasing line.
On the interval $ [1, 2] $, $ h(x) = 2 - x $, which is a straight, decreasing line.

At $ x = 1 $, the two lines meet, and the function remains continuous.

Proof

To prove that $ h $ is continuous, we need to show that the preimage of any closed set in $ Y $ is closed in $ X $. This is because continuity can be defined through the preimage of closed sets.

Formally, we need to show that if $ C \subseteq Y $ is closed in $ Y $, then $ h^{-1}(C) $ is closed in $ X $.

Since $ h(x) $ is defined differently on $ A $ and $ B $, we can write the preimage of $ C $ as:

$$ h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C). $$

Where:

$ f^{-1}(C) $ is the set of points in $ A $ whose image under $ f $ belongs to $ C $.
$ g^{-1}(C) $ is the set of points in $ B $ whose image under $ g $ belongs to $ C $.

Since $ f $ is continuous and $ C $ is closed in $ Y $, we know that $ f^{-1}(C) $ is closed in $ A $.

Because $ A $ is closed in $ X $, $ f^{-1}(C) $, being a closed subset of $ A $, is also closed in $ X $.

Similarly, since $ g $ is continuous and $ C $ is closed in $ Y $, $ g^{-1}(C) $ is closed in $ B $. And since $ B $ is closed in $ X $, $ g^{-1}(C) $ is closed in $ X $.

Now, $ h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C) $ is the union of two closed sets in $ X $, and the union of closed sets is itself closed in a topological space. Therefore, $ h^{-1}(C) $ is closed in $ X $.

Thus, we can conclude that the function $ h $ is continuous on $ X $. This completes the proof of the Pasting Lemma.

And so on.