Let $X$ be a topological space, and let $r : X \to A$ be a retract. Declare an equivalence relation $\sim$ on $X$ by saying $x \sim y$ iff $r(x) = r(y)$. Let $Y$ denote the resulting quotient space. From the UMP for quotients there exists a continuous map $h : Y \to A$ defined by $h([x]) = r(x)$. Let $k : A \to Y$ be the function defined by $k(a) = [a]$. If $k$ is continuous, then $Y \cong A$ since $k \circ h = 1_Y$ and $h \circ k = 1_A$. But is $k$ continuous?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
Here's how it works:
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
It's certainly continuous. Consider the inclusion map $\iota\colon A\to X$, and let $\pi \colon X\to Y$ be the quotient map. Then $k=\pi \circ \iota$ is the composition of two continuous maps and is therefore itself continuous.