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My question is part of an exercise in Hatcher's 'Algebraic Topology'.

Consider a CW complex $X$, constructed from a circle and two 2-disks $e_2$ and $e_3$, attached to that circle by maps of degree 2 and 3, respectively. Can someone show that $X$ is homotopy equivalent to a 2-sphere? Its 2-homology is generated by $3e_2 - 2e_3$, hence this homotopy equivalence $S^2 \to X$ must be at least 2-to-1 in a generic point.

It is easy to construct maps $S^2 \overset{f}{\to} X \overset{g}{\to} S^2$ that have degree one, hence compose to homotopy identity, but I am really stuck with $X \overset{g}{\to} S^2 \overset{f}{\to} X$... Is there a nice explanation why this should be homotopy identity?..

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    As it is, you're only looking for maps that evidence $S^2$ and $X$ as *homotopy retracts* of each other. It should be the same maps $f:S^2 \rightarrow X$ and $g:X \rightarrow S^2$ that have $fg \simeq 1_X$ and $gf\simeq 1_{S^2}$.2012-09-21
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    @AaronMazel-Gee Well, you are right, I'm a bit inaccurate, but if $X$ was really a homotopy sphere, this would not make any difference, since any degree-1 maps would do.2012-09-21
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    I'm facing the same problem. I believe the map f:S 2 →X is given so that the upper semisphere is mapped to e2, and lower semisphere is mapped to e3. But I cannot give the homotopy inverse of this map.2012-12-05
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    The crucial point is that when two attaching maps f,g:X->Y are homotopic, you will get two homotopic spaces which are obtained by attaching X to Y via f,g.2012-12-09
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    See Proposition 0.18 of Hatcher.2012-12-09
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    @lee: I couldn't figure out how this helps. :) However, an answer using homotopy theory would be that since both spaces are simply connected, any map $f$ that induces isomorphism on homology is a homotopy equivalence.2013-01-16

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