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?

  • $\begingroup$ 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

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 $g' \circ h = r \circ g \circ h = r \circ g \circ h \circ r \circ i = r \circ g \circ f \circ i\simeq r \circ id_X \circ i = r \circ i = id_A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.