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I'm trying exercise 9 on page 53 in Hatcher but I need some help with it. The exercise is:

In the surface $M_g$ of genus $g$, let $C$ be a circle that separates $M_g$ into two compact subsurfaces $M_h^\prime $ and $M_k^\prime$ obtained from the closed surfaces $M_h$ and $M_k$ by deleting an open disk from each. Show that $M_h^\prime$ does not retract onto its boundary circle $C$, and hence $M_g$ does not retract onto $C$. [Hint: abelianize $\pi_1$.] But show that $M_g$ does retract onto the nonseparating circle $C^\prime$ in the figure.

enter image description here

My first question is: assume there wasn't the hint, how would I think of abelianising? What does it mean exactly?

I thought I could do this by contradiction: if it retracts the induced map $i_\ast : \pi_1(C) \rightarrow \pi_1(M_h^\prime) $ is injective. I know $\pi_1(C) \cong \mathbb{Z}$, then I computed $\pi_1(M_h^\prime) \cong \mathbb{Z}$ and then I think I'm stuck. Right? Do you agree with $\pi_1(M_h^\prime) \cong \mathbb{Z}$ and being stuck after that?

What do I need to know to make progress? Many thanks for your help!

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    $r\colon M_g\rightarrow C$, $(g\geq 1)$ will be retraction if it is continuous and its restriction to $C$ is identity. But just regarding $r$ as continuous map, it will induce ***injection*** $r_{*}\colon \pi_1(M_g)\rightarrow \pi_1(C)$; and $\pi_1(M_g)$ is free abelian group of rank atleast $2$, and $\pi_1(C)\cong \mathbb{Z}$.2011-09-02
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    @group: Thank you! That is pretty neat! So the abelianization is only needed to prove the second part of the exercise?2011-09-02
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    @group Sorry, how do you get an injection?2011-09-02
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    @Dylan Moreland $r\circ i=Id_C$, so $r_*\circ i_*=Id$.2011-09-02
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    So from that I get that $i_*$ is injective and $r_*$ is surjective.2011-09-02
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    I also don't see how you're getting $\pi_1(M_h') = \mathbf Z$.2011-09-02
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    Oops. I glanced at Matt's post and saw that $i_*$ was an injection and didn't pay attention to the fact that group said $r_*$ was the injection.2011-09-02
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    Ah, you're different people!2011-09-02
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    Wouldn't it also work to use the fact that $H_1(Mg,\mathbb Z) =\mathbb Z^{2g} $, but $H_1(C,\mathbb Z)=\mathbb Z$, and get a contradiction with the induced maps composed with the retraction?2011-09-02
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    I am different people too.2011-09-02
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    This is a contradiction $r_{*}\colon \pi_1(M_g)\rightarrow \pi_1(C)$ be injective?2012-12-08

2 Answers 2

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I think the Wikipedia article on commutators explains the definitions and properties well. If I can clarify anything said there, let me know. It's important to note that abelianization is a covariant functor, so in particular a homomorphism of groups induces one of the abelianizations.

This is a useful trick, not only because in applying $\pi_1$ you will often produce free products that seem hard to compare with one another. For example, $\mathbf Z * \mathbf Z$ contains free subgroups of all finite ranks, so it isn't clear (at least to me) that we can use van Kampen to show that the wedge of two circles and the wedge of three circles have non-isomorphic $\pi_1$s. But abelianizing makes this clear.

Back to the example. I claim that for a retract $M_h' \to C$ to exist, $i_*^\text{ab}\colon \pi_1(C) \to \pi_1(M_h')^\text{ab}$ would have to be injective (It does not always happen that the abelianization of an injection is injective; what's special here?). Now, $M_h'$ is the surface of genus $h$ with a hole in it, and we can represent that as

enter image description here

So the map $i_*$ sends a generator for $\pi_1(C)$ to a commutator $[a_1, b_1] \cdots [a_g, b_g]$. Why does this contradict the injectivity of $i_*^\text{ab}$?

For the second part, you'll need to give a map. Here's one idea: in the above picture, ignoring the $C$, try to retract onto $a_1$. It might help to redraw this as a square with sides $a_1, b_1, a_1$, and a bunch of stuff coming off of the fourth side. The square looks a lot like the torus, and you know how to retract the torus onto a meridian (?). Note that you really can crush a bunch of stuff on the polygon together as long as you respect the identifications.

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    Woops, that $b_g$ should be a $b_h$ in the picture and I don't want to make another trip to the scanner.2011-09-02
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    Can you please explain why $i^{ab}_*$ is injective?2017-07-06
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Here is an approach using homology to show that $Mg$ does not retract to $C$:

Use the fact (proven here: Homology of surface of genus $g$), that $H_1(Mg, \mathbb Z) =\mathbb Z^{2g}$, and that $H_1(C,\mathbb Z)=\mathbb Z$ . Then we use the standard method of composing the retraction with the inclusion:

If there was a retraction $r: Mg\rightarrow C $, then the composition $roi$ with the inclusion $i$ would give an isomorphism h:$\mathbb Z \rightarrow \mathbb Z$, which cannot happen, because of the intermediate group $\mathbb Z^{2g}$.

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    Sure; the composition $ior(C)=C$, gives us an isomorphism in homology from $\mathbb Z$ to itself, since $H_1(C,\mathbb Z)=\mathbb Z$, but the homology of $Mg$ is $\mathbb Z^{2g}$; I may have the composition in the wrong order; I think we need $i: C\rightarrow Mg \ r:Mg \rightarrow C $, so that we need a composition of group maps such that $\mathbb Z \rightarrow \mathbb Z^{2g} \rightarrow \mathbb Z$ is an isomorphism of $\mathbb Z$ to itself.2011-09-02
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    I'm slightly worried: what if, in that sequence, we had $n \mapsto (n, 0, \ldots, 0)$ and $(m_1, \ldots, m_{2g}) \mapsto m_1$ for the maps? It seems like you need some exactness in the middle. Sorry for my slowness.2011-09-02
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    But we can also use that; $\mathbb Z$ and $\mathbb Z^{2g} $ are not isomorphic , e.g., by the fundamental thm. of finitely-generated Abelian groups; a retraction would induce an isomorphism between $\mathbb Z$ and $\mathbb Z^{2g}$2011-09-02
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    Well, I was just thinking that the existence of a retraction $r:Mg \rightarrow C $ alone would lead to $H_1(Mg;\mathbb Z)$ being isomorphic to $H_1(C;\mathbb Z)$; by the fundamental thm. of fin. gen. Abelian groups, a group G has a unique representation as a sum $n\mathbb Z(+)T$ ; where T is torsion, so that $\mathbb Z$ cannot be isomorphic to $\mathbb Z^{2g}$ So algebraically, a retraction is not possible; I do agree that a non-isomorphism can induce the identity in homology, but, algebraically, a retraction is not possible.2011-09-02
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    Hum. I don't see why it has to induce an isomorphism. Certainly if we had a _deformation_ retraction, then we'd have isomorphisms all over the place.2011-09-02
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    Just to be sure I understand your objection: do you agree that if there was a retraction , the homology/homotopy groups would be isomorphic, and $\mathbb Z$ and $\mathbb Z^{2g}$ are not isomorphic.2011-09-02
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    I agree that those two groups are not isomorphic! But in the end of the problem we are going to construct a retract $M_g \to C'$, and there's no way that's an isomorphism on $H^1$.2011-09-02
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    But if there is a retraction r: X-->A , then A and X are homotopy-equivalent, aren't they? Then their respective groups would be isomorphic, right? My post addresses only the fact that there is no retraction between $Mg$ and $C$; I did not try to prove that there is a retraction between $M_g$ and $C'$2011-09-02
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    I have the impression that you believe I tried to prove something which I never did try to prove; I only (tried to) prove(d) that $M_g$ does not retract to $C$; nothing more than that, nothing re $C'$; I don't see how wedges of $S^1$ relate to my answer. Let some other tough guy downvote me anonymously, tho.2011-09-02
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    I understand that you weren't trying to prove the second part! But I don't see anything in your proof that you couldn't say about $C'$. I can drop this, if you like :)2011-09-02
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    Dear Gary, I agree with @Dylan. I don't see where you use in your arguments that $C$ is separating; as Dylan says, I don't see why your argument wouldn't equally well apply to $C'$ (to which is *shouldn't* apply). Somewhere you have to use that $C$ is separating. Regards,2011-09-02
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    Fair-enough, Matt ; I was just not clear on what Dylan's objection was until you both explained it in the last two posts, but you're (both) right; I'll look into it.2011-09-02
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    @gary Sorry about the confusion. Let me try to clarify the example: fix a point $x \in S^1$; take $X = \{0, 1\} \times S^1$ and identify $(0, x) \sim (1, x)$ to get the wedge of two circles $Y$, and let $q\colon X \to Y$ be the quotient map. Define $p\colon X \to S^1$, $(v, y) \mapsto y$. By the universal property ($p(0, x) = p(1, x)$), there is a $p_*\colon Y \to S^1$ such that $p = p_* \circ q$. I think this gives a retraction. If we take abelianizations, we get $\mathbf Z \to \mathbf Z \oplus \mathbf Z \to \mathbf Z$ and everything seems okay.2011-09-02