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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?

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up vote 3 down vote accepted

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.

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