Adjunction space

Remarks

The composite $X \to X \sqcup Y \to X \cup_f Y$ is a closed topological embedding, and $Y \setminus A$ maps homeomorphically onto an open subset. Many authors only ask that $A$ be a subspace; the closedness of $A$ is needed for these two facts.

Example: let $Y = D^2$ be the closed unit disc, $A = S^1$ its boundary circle, and $X = S^1$ with $f(z) = z^2$. Then $f$ is continuous and $A$ is closed in $Y$, so $X \cup_f D^2$ is an adjunction space. It is homeomorphic to the real projective plane.

Example: if $X$ is a one-point space, then $X \cup_f Y$ is the quotient $Y / A$ that collapses $A$ to a point.

Non-example: let $Y = [0, 1]$, let $A = (0, 1]$, let $X$ be a one-point space, and let $f$ be the constant map. The clause that $A$ is closed fails, and the quotient is a two-point space whose open sets are $\varnothing$, $\set{[A]}$ and the whole space. There the image of $X$ is the open point $[A]$, which is not closed.