Topologies properties of a closed disc with arcs on the boundary identified Fix a natural number $n$.
Let $D^2$ denote the unit disc in $\mathbb{R}^2$ and $S^1$ the boundary circle, and consider the quotient space $X=D^2/ \sim$ where we have the quotient on the boundary $$(\cos \theta, \sin \theta) \sim \left(\cos\left(\theta + \frac{2r\pi}{n}\right), \sin\left(\theta + \frac{2r\pi}{n}\right)\right)$$
In other words, we are identifying points on arc of length $\frac{2\pi}{n}$. Denote by $Y$ the image of $S^1$ under the quotient map.
Two questions. First, I'd like to prove that the inclusion $Y \hookrightarrow X \setminus \{(0,0)\}$ is a deformation retract. I'd also like to compute the fundamental group $\pi_{1}(X, x_0)$ for some basepoint $x_0 \in X \setminus Y$.
Some thoughts. I don't think a simple deformation retract such as $((x,y),t) \mapsto (x,y)(1-t) + t\frac{(x,y)}{|(x,y)|}$ works because we have continuity issues where the angle of $(x,y)$ (in polar coordinates) is close to $\frac{2r\pi}{n}$ for some integer $r$. I've tried constructing more complicate retracts, but with little progress.
 A: Let us first understand the identification of points.
We regard $S^1$ as subspace of $\mathbb C$. Then the equivalence relation $\sim $ means $z \sim \zeta^k_n z$ where $\zeta_n = e^{2\pi i/n}$ is the $n$-th root of unity with smallest positive argument. The set of the $\zeta^k_n$ is the set of all $n$-th roots of unity.
In other word, we have $z \sim z'$ if and only if $z^n = (z')^n$. The "only if" part is obvious. For the converse note that $z^n = (z')^n$ means $(z/z')^n = 1$, i.e. $z/z'$ is an $n$-th root of unity.
We conclude that $X = D^2/\sim$ is nothing else than the adjunction space $C^2(n) = D^2 \cup_{\mu_n} S^1$, where $D^2$ is attached to $S^1$ by $\mu_n : S^1 \to S^1,\mu_n(z) = z^n$. In fact, let $D^2 + S^1$ be disjoint union, $p : D^2 + S^1 \to D^2(n)$ be the quotient map and $i : D^2 \to D^2 + S^1, j : S^1 \to D^2 + S^1$ be the canonical embeddings of the two summands. Since $\mu_n$ is surjective, we have $p(D^2) = C^2(n)$. The map $q = p \circ i : D^2 \to C^2(n)$ is a surjection and $j' = p \circ j : S^1 \to C^2(n)$ is an embedding. $C^2(n)$ is obtained from the disjoint union $D^2 + S^1$ by identifying $z \in S^1 \subset D^2$ with $z^n \in S^1$ in the second summand. This means that $z,z' \in D^2$ are sent to the same point in  $C^2(n)$ iff $z, z' \in S^1$ and $z^n = (z')^n$. Thus $q(z) = q(z')$ iff $z \sim z'$. Hence $q$ induces a continuos bijection $ \bar q : X \to C^2(n)$. Since both space are compact Hausdorff, $\bar q$ is a homeomorphism. It also shows that $\bar q$ identifies $Y$ with $j'(S^1) \approx S^1$.
$C^2(n)$ is the mapping cone of $\mu_n$. This can be regarded as the mapping cylinder $Z(\mu_n)$ of $\mu_n$ with the cone $CS^1 = D^2$ attached to the top of $Z(\mu_n)$ (i.e. the points of the base $S^1 \times \{0\}$ of $CS^1$ are identified with the point of the top $S^1\times \{1\}$ of $Z(\mu_n)$). Thus removing $0$ from $D^2/\sim$ is the same as removing the tip of $CS^1$ from $Z(\mu_n) \cup CS^1$. The resulting space deformation retracts to $Z(\mu_n)$ and the latter deformation retracts to the base $S^1$ of $Z(\mu_n)$. This answers your first question. If you consider about the above constructions, you also see that your suggestion of a (strong) deformation retraction works in fact. To see this directly, note that you actually defined a map
$$H : (D^2 \setminus \{0\} )\times I \to D^2 \setminus \{0\} , H(z,t) = (1-t)z + t \frac{z}{\lvert z \rvert}$$
which is clearly a strong deformation retraction onto $S^1$. If $z \sim z'$ with $z \ne z'$, we must have $z, z' \in S^1$, hence $H(z,t) = z \sim z' = H(z',t)$ for all $t$. Going to the quotients, we see that $H$ induces a strong deformation retraction of $X \setminus \{0\} $ onto $Y \approx S^1$.
Your second question can answered by the Seifert-van Kampen theorem. We have $C^2(n) = A \cup B$ with $A  = C^2(n) \setminus \{\text{tip of } CS^1\}, B = C^2(n) \setminus j'(S^1)$, where $A \simeq S^1$, $A \cap B \approx S^1 \times (0,1) \simeq S^1$ and $B \approx D^2 \setminus S^1$. We get
$$\pi_1(C^2(n)) \approx (\pi_1(A) * \pi_1(B)) / N$$
where $N$ is the normal subgroup generated by all words of the form $i_A(g)i_B(g)^{-1}$ with $g \in \pi_1(A \cap B)$ and $i_A: \pi_1(A \cap B) \to \pi_1(A)$ and $i_B: \pi_1(A \cap B) \to \pi_1(B)$ induced by the inclusions $\iota_A: A \cap B \to A, \iota_B: A \cap B \to B$. But $\pi_1(B) = 0$ and thus all  $i_B(g)^{-1} = 0$. Moreover $\pi_1(A) \approx \mathbb Z \approx \pi_1(A \cap B)$. We conclude that
$$\pi_1(C^2(n)) \approx \pi_1(A)/G ,$$
where $G$ is the (normal) subgroup generated by all $i_A(g)$ with $g \in \pi_1(A \cap B)$. Since $\pi_1(A) \approx \mathbb Z$, $G$ is nothing but the image of $i_A$. If we identity $\pi_1(A)$ with $\mathbb Z$, then $G  = k \mathbb Z$ for some $k$. Hence $\pi_1(C^2(n)) \approx \mathbb Z_k$. Let us show that $k = n$.
In fact we have a commutative diagram
$\require{AMScd}$
\begin{CD}
\pi_1(S^1) @>{f_*}>> \pi_1(Z(\mu_n))  @>{}>> \pi_1(A) \\
@A{(\mu_n)_*}AA @A{}AA @A{i_A}AA \\
\pi_1(S^1) @>{g_*}>> \pi_1(S^1 \times \{1\}) @>{}>> \pi_1(A \cap B) \end{CD}
where all unnamed arrows are the inclusion-induced homomorphism, $f: S^1 \to Z(\mu_n)$ embeds of $S^1$ as the base of the mapping cylinder and $g$ is the obvious homeomorphism. Note that all horizontal arrows are isomorphisms.
But is clear that the image of $(\mu_n)_*$ is generated by $(\mu_n)_*([id]) = [\mu_n] = n [id]$.
