[proofplan] The argument parallels the [high-dimensional cell attachment](/theorems/1906): we cover $Y := X \cup_f D^2$ by $A = Y \setminus \{0\}$ (which deformation retracts onto $X$) and $B = \mathring{D}^2$ (contractible), with $A \cap B \simeq S^1$. The decisive difference is that $\pi_1(S^1) \cong \mathbb{Z}$ is now non-trivial, so the amalgamation from [Seifert–van Kampen](/theorems/1905) contributes a single relation: the generator of $\pi_1(A \cap B)$ must equal the identity in $\pi_1(B) = \{e\}$. Tracing this generator through the homotopy equivalence $A \simeq X$ identifies it with $[f] \in \pi_1(X, x_0)$. The resulting amalgamated product collapses to the quotient $\pi_1(X, x_0) / \langle\langle [f] \rangle\rangle$. [/proofplan] [step:Construct the open cover $\{A, B\}$ and identify the three homotopy types] Write $Y := X \cup_f D^2$ and let $q: X \sqcup D^2 \to Y$ be the quotient map, so $q(a) = q(f(a))$ for $a \in S^1 = \partial D^2$. Fix $0 \in \mathring{D}^2$ and define \begin{align*} A &:= Y \setminus \{q(0)\} = X \cup_f (D^2 \setminus \{0\}), \\ B &:= q(\mathring{D}^2) \cong \mathring{D}^2. \end{align*} Openness, coverage, and path-connectedness of $A, B, A \cap B$ follow exactly as in the proof of [Attaching High-Dimensional Cells](/theorems/1906): $\{q(0)\}$ is closed, $B$ is the $q$-image of the open subset $\mathring{D}^2$, and $A \cap B = \mathring{D}^2 \setminus \{0\}$. The three homotopy equivalences are: - $B \simeq \{*\}$ via the straight-line contraction to $0$, so $\pi_1(B) = \{e\}$. - $A \simeq X$ via radial deformation retraction of the punctured disk onto $S^1$ composed with the identification along $f$. - $A \cap B = \mathring{D}^2 \setminus \{0\} \simeq S^1$ via radial projection onto a circle $\{|x| = r\}$ with $0 < r < 1$. In particular, $\pi_1(A \cap B, y_1) \cong \pi_1(S^1, *) \cong \mathbb{Z}$ by the [Fundamental Group of the Circle](/theorems/1894), where $y_1 \in A \cap B$ is any choice of basepoint. [/step] [step:Identify the image of the generator of $\pi_1(A \cap B)$ in $\pi_1(A)$] Fix $y_1 = q(r e_1) \in A \cap B$ with $0 < r < 1$ (so $y_1$ lies on a small circle of radius $r$ around the puncture). Let $c_r: [0,1] \to A \cap B$ be the loop $c_r(t) := q(r e^{2\pi i t})$ that traverses this small circle once counterclockwise. By the [Fundamental Group of the Circle](/theorems/1894), $[c_r]$ generates $\pi_1(A \cap B, y_1) \cong \mathbb{Z}$. Let $i_A: A \cap B \hookrightarrow A$ denote the inclusion. We compute the class $(i_A)_*[c_r] \in \pi_1(A, y_1)$. [claim:The class $(i_A)_*[c_r] \in \pi_1(A, y_1)$ corresponds to $[f]$ under the isomorphism $\pi_1(A, y_1) \cong \pi_1(X, x_0)$ determined by the retraction and a basepoint-change path] [proof] The radial homotopy $(x, t) \mapsto (1-t) x + t x/|x|$ on $D^2 \setminus \{0\}$ is a deformation retraction of the punctured disk onto $S^1$. Passing through the quotient $q$, this yields a deformation retraction $r: A \to X$ whose restriction to $A \cap B$ is the radial projection $q(s e^{i\theta}) \mapsto q(e^{i\theta}) = f(e^{i\theta})$. Apply this retraction to the loop $c_r$: the image of $c_r$ in $X$ under $r$ is \begin{align*} r(c_r(t)) = r(q(r e^{2\pi i t})) = q(e^{2\pi i t}) = f(e^{2\pi i t}), \end{align*} which is exactly the loop $f: S^1 \to X$ traversed once. Since $r$ is a homotopy equivalence, $r_* [c_r] = [f] \in \pi_1(X, r(y_1))$ (after using a [change of basepoint](/theorems/1880) to compare $r(y_1) = f(e_1)$ to the theorem's chosen basepoint $x_0 = f(e_1)$; without loss of generality we may take $x_0 = f(e_1)$, in which case no basepoint adjustment is needed). Equivalently, the inclusion $i_A: A \cap B \hookrightarrow A$ composed with the deformation retraction $r: A \to X$ sends $[c_r]$ to $[f]$. Under the isomorphism $\pi_1(A, y_1) \cong \pi_1(X, x_0)$ induced by $r$ and the basepoint-change path, $(i_A)_* [c_r]$ corresponds to $[f]$. [/proof] [/claim] Let $i_B: A \cap B \hookrightarrow B$ denote the other inclusion. Since $\pi_1(B) = \{e\}$, $(i_B)_*[c_r] = e$. [guided] **What loop generates $\pi_1(A \cap B)$?** The intersection $A \cap B = \mathring{D}^2 \setminus \{0\}$ deformation retracts onto a small circle around the puncture. By the [Fundamental Group of the Circle](/theorems/1894), $\pi_1$ of the small circle is $\mathbb{Z}$, generated by the once-around loop $c_r$. **Where does $c_r$ go in $A$?** The subspace $A$ contains the full punctured disk (after gluing), and the punctured disk deformation retracts onto its boundary $S^1$ via radial projection $x \mapsto x/|x|$. Under this retraction, the small circle $\{|x| = r\}$ maps onto the boundary circle $\{|x| = 1\} = S^1$ by scaling, i.e., $r e^{i\theta} \mapsto e^{i\theta}$. Composing with $q$ (which identifies the boundary $S^1$ with its image under the attaching map $f$), the small circle in $A \cap B$ retracts in $A$ to the loop $f: S^1 \to X \hookrightarrow A$. **So what is $(i_A)_*[c_r]$?** In $\pi_1(A, y_1)$, the loop $c_r$ is homotopic (via the retraction) to the attaching loop $f$. Translating to $\pi_1(X, x_0)$ via the homotopy equivalence $A \simeq X$, we get $(i_A)_*[c_r] \leftrightarrow [f]$. **And $(i_B)_*[c_r]$?** The subspace $B$ is contractible, so its fundamental group is trivial, and every loop — including $c_r$ — becomes the identity. **The stage is set for amalgamation.** The generator of the amalgamating group $\pi_1(A \cap B) = \mathbb{Z}$ maps to $[f]$ on one side and to $e$ on the other. The amalgamation relation in the pushout forces these two images to be equal — forcing $[f] = e$ in the target. This is precisely the relation imposing $[f] \in \langle\langle [f] \rangle\rangle$, i.e., killing the normal closure of $[f]$ in $\pi_1(X, x_0)$. [/guided] [/step] [step:Apply Seifert–van Kampen and identify the result with $\pi_1(X, x_0) / \langle\langle [f] \rangle\rangle$] The hypotheses of [Seifert–van Kampen](/theorems/1905) are verified in Step 1: $Y = A \cup B$, both open, and all three of $A, B, A \cap B$ path-connected (each is homotopy equivalent to a path-connected space). The theorem gives \begin{align*} \pi_1(Y, y_1) \cong \pi_1(A, y_1) *_{\pi_1(A \cap B, y_1)} \pi_1(B, y_1). \end{align*} Substituting $\pi_1(B, y_1) = \{e\}$ and $\pi_1(A \cap B, y_1) \cong \mathbb{Z}$ with generator $[c_r]$, \begin{align*} \pi_1(Y, y_1) \cong \pi_1(A, y_1) *_{\mathbb{Z}} \{e\}. \end{align*} We evaluate the right-hand side. The [Universal Property of the Free Product with Amalgamation](/theorems/1904) with $K = \pi_1(A, y_1) / \langle\langle (i_A)_* [c_r] \rangle\rangle$ shows that $\pi_1(A, y_1) *_\mathbb{Z} \{e\}$ is precisely this quotient: [claim:For any group $G$ with a homomorphism $\psi: \mathbb{Z} \to G$, the amalgamation $G *_\mathbb{Z} \{e\}$ (over $\psi$ on one side and the trivial map on the other) is isomorphic to $G / \langle\langle \psi(1) \rangle\rangle$] [proof] Let $N = \langle\langle \psi(1) \rangle\rangle \trianglelefteq G$ and $K = G/N$. Let $p: G \to K$ be the quotient and $q: \{e\} \to K$ the trivial map. The compatibility $p \circ \psi = q \circ 0$ holds: $p(\psi(1)) = 0$ in $K$ since $\psi(1) \in N = \ker p$, and by multiplicativity $p \circ \psi \equiv 0$; similarly $q \circ 0 \equiv 0$. By the universal property of $G *_\mathbb{Z} \{e\}$, there is a unique homomorphism $\Psi: G *_\mathbb{Z} \{e\} \to K$ with $\Psi \circ j_G = p$. Conversely, the inclusion $G \hookrightarrow G *_\mathbb{Z} \{e\}$ as $j_G$ sends $\psi(1)$ to $j_G(\psi(1)) = j_{\{e\}}(0) = e$ (using the amalgamation relation). Hence $N \subseteq \ker j_G$, so $j_G$ descends to $\bar j_G: K = G/N \to G *_\mathbb{Z} \{e\}$. The compositions $\bar j_G \circ \Psi$ and $\Psi \circ \bar j_G$ agree with the identity on the generating sets $j_G(G)$ and $p(G)$ respectively, hence are identities. Thus $G *_\mathbb{Z} \{e\} \cong G/N$. [/proof] [/claim] Applying the claim with $G = \pi_1(A, y_1)$ and $\psi = (i_A)_*$, where $\psi(1) = (i_A)_*[c_r]$: \begin{align*} \pi_1(A, y_1) *_\mathbb{Z} \{e\} \cong \pi_1(A, y_1) / \langle\langle (i_A)_*[c_r] \rangle\rangle. \end{align*} By Step 2, the homotopy equivalence $A \simeq X$ together with a [change of basepoint](/theorems/1880) identifies $\pi_1(A, y_1) \cong \pi_1(X, x_0)$ and sends $(i_A)_*[c_r]$ to $[f]$. Normal closures transport under isomorphisms, so \begin{align*} \pi_1(A, y_1) / \langle\langle (i_A)_*[c_r] \rangle\rangle \cong \pi_1(X, x_0) / \langle\langle [f] \rangle\rangle. \end{align*} Finally, using [change of basepoint](/theorems/1880) to pass from $y_1$ to $x_0$ on $Y$: \begin{align*} \pi_1(Y, x_0) \cong \pi_1(Y, y_1) \cong \pi_1(X, x_0) / \langle\langle [f] \rangle\rangle. \end{align*} [/step] [step:Identify the surjection as induced by the inclusion $X \hookrightarrow Y$ with kernel $\langle\langle [f] \rangle\rangle$] Let $\iota: X \hookrightarrow Y = X \cup_f D^2$ be the inclusion. We show the induced homomorphism \begin{align*} \iota_*: \pi_1(X, x_0) \to \pi_1(Y, x_0) \end{align*} is the quotient map $\pi_1(X, x_0) \twoheadrightarrow \pi_1(X, x_0)/\langle\langle [f] \rangle\rangle$ under the isomorphism of Step 3. The composition $X \hookrightarrow A \simeq X$ of the inclusion and the deformation retraction from Step 1 is homotopic to the identity on $X$, so on $\pi_1$ it equals the identity of $\pi_1(X, x_0)$. Hence the inclusion $\pi_1(X, x_0) \hookrightarrow \pi_1(A, x_0)$ (after basepoint-matching) is an isomorphism whose inverse is induced by the retraction $A \to X$. The canonical inclusion $k_A: \pi_1(A, y_1) \hookrightarrow \pi_1(A, y_1) *_\mathbb{Z} \{e\}$ becomes, under the claim in Step 3, the quotient $\pi_1(A, y_1) \to \pi_1(A, y_1)/\langle\langle (i_A)_*[c_r] \rangle\rangle$. Composing: \begin{align*} \pi_1(X, x_0) \xrightarrow{\;\cong\;} \pi_1(A, y_1) \twoheadrightarrow \pi_1(A, y_1)/\langle\langle (i_A)_*[c_r] \rangle\rangle \xrightarrow{\;\cong\;} \pi_1(Y, x_0). \end{align*} The first arrow is the isomorphism from the $A \simeq X$ retraction with basepoint change; the middle is the quotient map; the third is the Seifert–van Kampen isomorphism. This composite equals $\iota_*$ (by naturality: each step is the induced map of an inclusion, deformation-retract, or quotient). Hence $\iota_*: \pi_1(X, x_0) \to \pi_1(Y, x_0)$ is the quotient map with kernel $\langle\langle [f] \rangle\rangle$, proving \begin{align*} \pi_1(Y, x_0) \cong \pi_1(X, x_0) / \langle\langle [f] \rangle\rangle \end{align*} as asserted. [/step]