Proof of the Adhesive Lemma
Theorem
For a topological space $X,Y$, suppose two closed sets $A,B \subset X$ satisfy $A \cup B = X$ and two continuous functions $f : A \to Y$ and $g : B \to Y$ satisfy $f(x) = g(x)$ for every $x \in A \cap B$. Then the $h$ defined as follows is a continuous function. $$ h(x) : = \begin{cases} f(x), & x \in A \\ g(x), & x \in B \end{cases} $$
Explanation
This lemma, also called the “full gluing lemma”, is so straightforward that one can understand it merely by reading the statement. The reason it has been given the name “lemma” and even a proof is probably that it is used so frequently that naming it is convenient and, moreover, it is provable.
Proof
To show that $h$ is continuous, consider a closed set $C \subset Y$.
Since $f$ and $g$ are continuous, $f^{-1}(C)$ is a closed set in $A$ and $g^{-1}(C)$ is a closed set in $B$. By the assumption that $A$ and $B$ are closed in $X$, it follows that $f^{-1}(C)$ and $g^{-1}(C)$ are also closed in $X$. $$ \begin{align*} h^{-1} (C) =& \left\{ x \in X \ | \ h(x) \in C \right\} \\ =& \left\{ x \in A \ | \ h(x) \in C \right\} \cup \left\{ x \in B \ | \ h(x) \in C \right\} \\ =& \left\{ x \in A \ | \ f(x) \in C \right\} \cup \left\{ x \in B \ | \ g(x) \in C \right\} \\ =& f^{-1} (C) \cup g^{-1} (C) \end{align*} $$ Because this is a union of closed sets in $X$, $h^{-1}(C)$ is closed in $X$, and $h$ is a continuous function.
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Definition
Let the paths $p,q : [0,1] \to X$ and $p(1) = q(0)$ satisfy $p(1) = q(0)$. Then define $p*q : [0,1] \to X$ as follows and call $p,q$ the product path. $$ (p \ast\ q) (t) = \begin{cases} p(2t), & 0 \le t \le 1/2 \\ q(2t-1), & 1/2 \le t \le 1 \end{cases} $$
The product path can be viewed as the concept obtained by applying the gluing lemma at exactly one point, and it is useful in the study of path-connectedness.
Homotopy
Gluing paths in this way leads into the study of homotopy in algebraic topology, and one encounters the gluing lemma itself so frequently that it becomes tiresome.