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Let me write definition of Pasting Lemma: Let $X=A \cup B$ where $A$ and $B$ are closed subsets of $X$. Let $g: A\to Y$ and $h:B\to Y$ be continuous function such that $g|A\cap B = h|A\cap B$. Define $f:X\to Y$ by $f\left( x\right) =\begin{cases}g\left( x\right) ,x\in A\\ h\left( x\right) ,x\in B\end{cases}$

then $f$ is continuous.

My request is can you summary the pasting lemma for me?

I looked some applications of pasting lemma and author says that $f$ is continuous since $g$ and $h$ are continuous. Does theorem say that like this? Is it enough to check $g$ and $h$ are continuous? Thank all of you for now.

Fuat Ray
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    I know this may sound complicated, but this is quite intuitive. The theorem says that if you have two continuous functions which coincides in the interesction of its domains then it can be extended to the union. Do you undertand the proof, what do you need exactly? If you study homotopy theory, you will see many aplications of this lemma, which can help you to undestand it. – Marcos Jan 15 '22 at 22:22
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    For applications, it is not enough to check that $f$ and $g$ are continuous, because there are other hypotheses you must also check: You must check that $A$ and $B$ are closed; You must also check that $g \mid A \cap B = f \mid A \cap B$. – Lee Mosher Jan 15 '22 at 22:48

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I think the best way to understand the pasting lemma is to understand why a "naive" version of it is false:

(Incorrect "theorem") Suppose $A,B$ are subspaces of a topological space $X$, and $f:A\rightarrow Y$, $g:B\rightarrow Y$ are continuous functions which agree on the intersection of their domains (that is, for $x\in A\cap B$ we have $f(x)=g(x)$). Then the "union" of $f$ and $g$ $$A\cup B\rightarrow Y: x\mapsto\begin{cases} f(x) & \mbox{ if } x\in A\\ g(x), & \mbox{ if } x\in B\\ \end{cases}$$ (which makes sense since $f$ and $g$ agree on $A\cap B$) is again continuous.

To see why this is false, consider for example $A=[0,1]$, $B=(1,2]$, $f:[0,1]\rightarrow\mathbb{R}: x\mapsto 0$, and $g: B\rightarrow\mathbb{R}: x\mapsto 1$ (where all spaces involved have the usual topologies). The union of $f$ and $g$ is the function $$[0,2]\rightarrow\mathbb{R}: x\mapsto\begin{cases} 0 & \mbox{ if } x\le 1\\ 1 & \mbox{ if } x>1\\ \end{cases}$$ which is not continuous.

So we have a naive idea that gluing together continuous functions should result in a continuous function, but a simple counterexample to the most obvious way of stating that. The correct pasting lemma is what we get when we realize what additional hypotheses we were missing in order to rule out counterexamples like the above - namely, that both $A$ and $B$ must be closed (this can actually be weakened somewhat, but that's not worth focusing on at first).

Noah Schweber
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