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Show that a retract of a contractible space is contractible. This exercise come from the Hatcher' book.

This is my proof.

Since $X$ is contractible, there are maps $f: X\rightarrow x$ and $g:x\rightarrow X.$Such that $f\circ g=id, g\circ f=id.$ i.e. $X\simeq x$. Where $x$ is a point. Let $r: X\rightarrow A$ be the retract.

Then define $h:A\rightarrow x$. And then $h\circ r\circ g=id_x$, $r\circ g\circ h=id_A$.

Is this correct?

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  • $\begingroup$ You defined $h$ and then immediately drawed a conlusion...... I think, for a proof, you need to add more details about how you conclude it (for instances, construct a homotopy between two maps) $\endgroup$ – Kevin. S Apr 18 '20 at 8:34
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Your proof has the correct idea, but has a gap. If $x$ denotes a one-point space, then certainly $f \circ g = id_x$, but $g \circ f = id_X$ is not true. You only have $g \circ f \simeq id_X$.

Let $i : A \to X$ denote inclusion and $g' = r \circ g$. We have $r \circ i = id_A$ and $h \circ r = f$. Therefore $h \circ g' = h \circ r \circ g = f \circ g = id_x$ and $g' \circ h = r \circ g \circ h = r \circ g \circ h \circ r \circ i = r \circ g \circ f \circ i\simeq r \circ id_X \circ i = r \circ i = id_A$.

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