Let $X$ and $Y$ be topological spaces, and $f:A\rightarrow Y$ a continuous map from a subset $A$ of $X$ to $Y$.
We can form the adjunction space $X \cup_f Y$ by appropriately quotienting the disjoint union $X \sqcup Y:= (X \times \{1\}) \cup (Y \times \{2\})$, where we define an equivalence relation on $X \sqcup Y$ by identifying every element $(y,2) \in f(Y) \times \{2\}$, with the elements $(a,1) \in A \times \{1\}$ such that $f(a) = y$ (and then taking the reflexive closure of this relation). The elements of $X \cup_f Y$ are then the equivalence classes of $(z,i) \in X \sqcup Y$ under this relation. These can be described explicitly as:
- $ [(x,1)] = \{ (x,1) \} $ for all $x \in X\backslash A$,
- $[(a,1)] = \{ (f(a),2) \} \cup \{ (a',1) \ | \ a' \in f^{-1}(f(a)) \}$ for all $a \in A$,
- $[(y,2)] = \{ (y,2) \}$ for all $y \in Y\backslash f(A)$, and
- $[(y,2)] = \{ (y,2) \} \cup \{(a,1) \ | \ f(a) = y \} $ for all $y\in f(A)$
Question: Almost everybody defines the adjunction space assuming that $A$ is a closed subset of $X$. Why is this necessary? It seems as though everything that I have described above is well-defined without using this assumption.
Imagine that you glue together two copies of $[0,1]$ using $A=(0,1]$ and the identity map.
Then each positive number corresponds to one point in the quotient ($(x,1)\equiv (x,2)$ for all $x>0$), but there are two zeroes ($(0,1)\not \equiv (0,2)$).
Therefore any sequence which would normally tend to $0$ now has two different limits! Of course this last statement depends on the topology or whatever means you use to define limits in the quotient space, but there are no natural ways to avoid weirdness here.