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Consider $N_g$, the non-orientable surface of genus $g$, i.e. the quotient space of a regular $2g-$gon, obtained by gluing pairs of adjacent sides in the $2g$-gon (as a reference: Non orientable surface of genus $g$).

Now consider $X$ to be the space $N_g$ without a point. Does $X$ deformation retract to a wedge of a number of circles, specifically, a wedge of $g$ circles?

The context of the question is the following: if we consider $\Sigma_h$, the orientable surface of genus $h$ (i.e. connected sum of $h$ tori) and we remove a point, then the resulting space deformation retracts to a wedge of $2h$ circles. Then, I wondered when is $\Sigma_h$ without a point homotopy equivalent to $N_g$ without a point, and the above "conjecture" answers the question.

If we consider $N_g$ with a point in the $2g-$gon model, then it deformation retracts to the boundary of the $2g-$gon, whose pairs of adjacent sides are glued together, but I am not sure if this is just the wedge of $g$ circles. If this is not the case, to which space does $N_g$ without a point deformation retract?

As a final remark, the fundamental group of $N_g$ without a point is isomorphic to the fundamental group of the wedge of $g$ circles.

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It does indeed, and the proof is exactly the same as in the orientable case.

Let $P$ be the $2g$-gon with its gluing pattern as you described, and let $\pi : P \to N_g$ be the quotient map. Let $\mathcal O$ be the point at the center of $P$. Then $X = N_g - \pi(\mathcal O)$.

Next, observe that the gluing pattern of $P$ has one vertex cycle. To put this another way, $\pi$ takes all vertices of $P$ to a single point. It follows $\pi(\partial P) \subset \pi(P)=N_g$, which is the image in $N_g$ of the boundary $\partial P$, is a wedge of $g$ circles in $N_g$, because $\pi(\partial P)$ is a 1-dimensional CW complex with one vertex and $g$ edges. Removing the point $\pi(\mathcal O)$, it follows that $\pi(\partial P)$ is a wedge of $g$ circles in $X$.

Now consider the retraction map $r : P - \mathcal O \to \partial P$ which takes each "radial segment" of $P - \mathcal O$ to the unique endpoint of that segment on $\partial P$. There is an obvious deformation retraction $$H : (P-\{\mathcal O\}) \times [0,1] \to P - \{\mathcal O\} $$ from the identity map of $P - \mathcal O$ to the retraction map $R$ defined as follows: if $x$ lies on a radial segment having endpoint $q \in \partial P$ then $H(x,t) = (1-t)x + t q$.

The final observation is that this deformation retraction on $P - \mathcal O$ respects all of the identifications made by the restricted quotient map $\pi : P - \mathcal O \to N - \pi(\mathcal O) = X$, and therefore it descends to a deformation retraction from $X$ to $\pi(\partial P)$.

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  • $\begingroup$ At the end of the third paragraph: you mean $\pi(\partial P)$ is the wedge of $g$ circles in $X$? $\endgroup$
    – Luke
    Mar 19 '20 at 15:02
  • $\begingroup$ Yes, thanks for the fix. $\endgroup$
    – Lee Mosher
    Mar 19 '20 at 15:12
  • $\begingroup$ Thank you for the answer! $\endgroup$
    – Luke
    Mar 20 '20 at 9:38

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