# Does the punctured non-orientable surface of genus $g$ deformation retract to a wedge of $g$ circles?

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.

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 $$\pi(\partial P)$$.

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