Adjunction Space , Collapsing boundary to a point I was studying Adjunction space from some lecture notes. I have the following Question, Suppose I take $X=D^{2}$, closed unit disk in plane. I take $A=S^{1}$ , the boundary of $D^2$. Let $Y$ denote the singleton set $\\{1\\}$. Let $f$, $$f:S^{1} \longrightarrow Y$$ be the attaching map, defined by $f(x)=1$, that is a constant map. How does Adjunction space $X\sqcup_{f}Y$ look like. What I understand is that we are collapsing the boundary to a point.

You are correct. By definition of the adjunction space, if $Y$ is a one-point space then $X \sqcup_f Y = X/A$ . In this case, the resulting space is a sphere, $D^2/S^1 = S^2$.

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Indeed, let $p$ denote a point in $S^2$. By the stereographic projection, there is a homeomorphism $\operatorname{int}(D^2) \to S^2 - p$ from the open disk to the sphere minus a point. This extends to a continuous map $D^2\to S^2$ by mapping the boundary points to $p$. Moreover, by the universal property of a quotient space, this map factors through $D^2/S^1$, i.e., there is an induced continuous map $D^2/S^1 \to S^2$. This is clearly a bijection from a compact space to a Hausdorff space, and hence it is a homeomorphism.