[proofplan] The proof uses the Morse-theoretic form of the Lefschetz hyperplane theorem: after choosing the hyperplane section smoothly, the pair $(X,Y)$ has the homotopy type of a relative CW-complex obtained from $Y$ by attaching cells of real dimension at least $n$. Once this cell-attachment statement is available, the conclusion is purely homotopy-theoretic. Cells of dimension at least $n$ make the relative homotopy groups $\pi_k(X,Y)$ vanish for $k \leq n-1$, and the long exact sequence of the pair then gives isomorphisms below degree $n-1$ and surjectivity in degree $n-1$. [/proofplan]

[step:Reduce the theorem to the Lefschetz cell-attachment form]We use the following standard Morse-theoretic Lefschetz input: for a smooth hyperplane section $Y = X \cap H$ of a smooth complex projective manifold $X$ of complex dimension $n$, there is a relative CW-pair $(Z,Y)$ and a homotopy equivalence of pairs under $Y$ \begin{align*} \Phi: (Z,Y) \longrightarrow (X,Y) \end{align*} such that $Z$ is obtained from $Y$ by attaching cells of real dimensions $d \geq n$. The phrase "under $Y$" means that if \begin{align*} j: Y \hookrightarrow Z \end{align*} denotes the CW-pair inclusion, then the composite $\Phi \circ j: Y \to X$ is homotopic to the original inclusion \begin{align*} i: Y \hookrightarrow X. \end{align*} This is the Morse-theoretic Lefschetz cell-attachment theorem for the pair $(X,Y)$. Its hypotheses apply here because $X \subset \mathbb P^N$ is a smooth complex projective manifold and $Y = X \cap H$ is a smooth hyperplane section. The Morse-theoretic proof uses a Lefschetz pencil or an equivalent strictly plurisubharmonic exhaustion adapted to the hyperplane section; the complex Hessian index estimate gives the handle decomposition of the pair $(X,Y)$ with relative handles of real index at least $n$. Thus the theorem supplies the cell-attachment conclusion needed here directly, not merely a CW-decomposition of the affine complement $X \setminus Y$. Since this result is not yet available as a resolved theorem in the wiki, we record the external dependency explicitly: (citing a result not yet in the wiki: Morse-theoretic Lefschetz cell-attachment theorem). Because homotopy groups of pairs are invariant under homotopy equivalence of pairs under $Y$, it suffices to prove the result for the relative CW-pair $(Z,Y)$ and then transfer it along $\Phi$.[/step]

[guided]The geometric content of the theorem is contained in the Lefschetz cell-attachment statement. It says that passing from the hyperplane section $Y$ to the whole manifold $X$ does not require attaching low-dimensional relative cells: all attached cells have real dimension at least $n$. We verify the hypotheses of this external input in the present setting. The theorem requires a smooth complex projective manifold and a smooth hyperplane section. These are exactly the assumptions in the statement: $X \subset \mathbb P^N$ is smooth and projective of complex dimension $n$, and $Y = X \cap H$ is assumed smooth. Therefore the Morse-theoretic Lefschetz cell-attachment theorem gives a CW-pair $(Z,Y)$ and a homotopy equivalence of pairs under $Y$ \begin{align*} \Phi: (Z,Y) \longrightarrow (X,Y), \end{align*} where $Z$ is constructed from $Y$ by attaching cells of real dimensions $d \geq n$. Under $Y$ means that, for the inclusion \begin{align*} j: Y \hookrightarrow Z, \end{align*} the composite $\Phi \circ j$ is homotopic to the original inclusion \begin{align*} i: Y \hookrightarrow X. \end{align*} The Morse-theoretic proof is a statement about the relative pair $(X,Y)$: one uses a Lefschetz pencil or an equivalent strictly plurisubharmonic exhaustion adapted to $Y$, and the complex Morse index estimate produces relative handles of real index at least $n$. This direction is important. A bound on the indices in the affine complement alone would not by itself prove the required lower bound on the dimensions of cells attached to $Y$; the external Lefschetz theorem supplies the relative handle conclusion directly. Homotopy equivalence of pairs preserves the relative homotopy groups $\pi_k(-,-)$ and is compatible with the long exact homotopy sequence of the pair. Therefore the desired statement for $(X,Y)$ follows once it is proved for $(Z,Y)$ and then transferred along the under-$Y$ equivalence $\Phi$. Since this input is not yet a resolved theorem in the wiki, the citation is recorded explicitly as an external dependency: (citing a result not yet in the wiki: Morse-theoretic Lefschetz cell-attachment theorem).[/guided]

[step:Show that the relative homotopy groups vanish below the attached cell dimension]Let $m$ be the smallest dimension of a cell attached to $Y$ in the CW-pair $(Z,Y)$. By the Lefschetz cell-attachment form, $m \geq n$. The relative CW-complex $(Z,Y)$ has no relative cells of dimension $d < m$. By the relative cellular approximation theorem and the relative homotopy computation for CW-pairs, a relative CW-pair with no cells of dimension less than $m$ satisfies \begin{align*} \pi_k(Z,Y,y_0) = 0 \end{align*} for every integer $k \leq m-1$ and every base point $y_0 \in Y$. Since $m \geq n$, this gives \begin{align*} \pi_k(Z,Y,y_0) = 0 \end{align*} for every $k \leq n-1$. Here we are using a standard CW-pair connectivity theorem: (citing a result not yet in the wiki: relative cellular approximation and connectivity of CW-pairs).[/step]

[guided]The point of the cell decomposition is that relative homotopy classes in low degrees cannot detect high-dimensional attached cells. Let $m$ denote the smallest dimension of any cell attached to $Y$ in order to obtain $Z$. The Lefschetz cell-attachment theorem gives $m \geq n$. A based element of $\pi_k(Z,Y,y_0)$ is represented by a continuous map of pairs \begin{align*} f: (D^k,S^{k-1}) \longrightarrow (Z,Y), \end{align*} where $D^k$ is the closed $k$-disk and $S^{k-1} = \partial D^k$. If $k \leq m-1$, the relative cellular approximation theorem allows $f$ to be homotoped relative to $S^{k-1}$ into the relative $(m-1)$-skeleton of $(Z,Y)$. But the relative $(m-1)$-skeleton is just $Y$, because no cells of dimension less than $m$ were attached. Hence $f$ is homotopic as a map of pairs to a map whose image lies in $Y$, so it represents the zero element of $\pi_k(Z,Y,y_0)$. Thus \begin{align*} \pi_k(Z,Y,y_0) = 0 \end{align*} for every $k \leq m-1$. Since $m \geq n$, we obtain \begin{align*} \pi_k(Z,Y,y_0) = 0 \end{align*} for every $k \leq n-1$. This step uses the standard CW-pair connectivity theorem, recorded as an external dependency because it is not yet resolved in the wiki: (citing a result not yet in the wiki: relative cellular approximation and connectivity of CW-pairs).[/guided]

[step:Apply the long exact homotopy sequence of the pair]For the based pair $(Z,Y,y_0)$, the long exact homotopy sequence gives, for each integer $k \geq 1$, an exact segment \begin{align*} \pi_{k+1}(Z,Y,y_0) \longrightarrow \pi_k(Y,y_0) \xrightarrow{j_*} \pi_k(Z,y_0) \longrightarrow \pi_k(Z,Y,y_0), \end{align*} where \begin{align*} j: Y \hookrightarrow Z \end{align*} is the inclusion map. If $1 \leq k < n-1$, then $k \leq n-2$, so both \begin{align*} \pi_{k+1}(Z,Y,y_0) = 0 \quad \text{and} \quad \pi_k(Z,Y,y_0) = 0 \end{align*} because $k+1 \leq n-1$ and $k \leq n-1$. Exactness then implies that \begin{align*} j_*: \pi_k(Y,y_0) \longrightarrow \pi_k(Z,y_0) \end{align*} is an isomorphism. For $k = n-1$, the exact segment \begin{align*} \pi_{n-1}(Y,y_0) \xrightarrow{j_*} \pi_{n-1}(Z,y_0) \longrightarrow \pi_{n-1}(Z,Y,y_0) \end{align*} and the vanishing \begin{align*} \pi_{n-1}(Z,Y,y_0) = 0 \end{align*} show that $j_*$ is surjective.[/step]

[guided]The long exact sequence of the pair turns relative vanishing into information about the inclusion-induced map on absolute homotopy groups. For the based inclusion \begin{align*} j: Y \hookrightarrow Z, \end{align*} the long exact homotopy sequence contains \begin{align*} \pi_{k+1}(Z,Y,y_0) \longrightarrow \pi_k(Y,y_0) \xrightarrow{j_*} \pi_k(Z,y_0) \longrightarrow \pi_k(Z,Y,y_0). \end{align*} First take $1 \leq k < n-1$. Then $k+1 \leq n-1$, so the previous step gives \begin{align*} \pi_{k+1}(Z,Y,y_0) = 0. \end{align*} It also gives \begin{align*} \pi_k(Z,Y,y_0) = 0. \end{align*} Exactness now says two things. Since the kernel of $j_*$ is the image of the zero group $\pi_{k+1}(Z,Y,y_0)$, the map $j_*$ is injective. Since the image of $j_*$ is the kernel of the map from $\pi_k(Z,y_0)$ to the zero group $\pi_k(Z,Y,y_0)$, the map $j_*$ is surjective. Hence $j_*$ is an isomorphism. Now take $k = n-1$. The relevant exact segment is \begin{align*} \pi_{n-1}(Y,y_0) \xrightarrow{j_*} \pi_{n-1}(Z,y_0) \longrightarrow \pi_{n-1}(Z,Y,y_0). \end{align*} The previous step gives \begin{align*} \pi_{n-1}(Z,Y,y_0) = 0. \end{align*} Therefore the image of $j_*$ is the kernel of the zero map out of $\pi_{n-1}(Z,y_0)$, which is all of $\pi_{n-1}(Z,y_0)$. Thus $j_*$ is surjective.[/guided]

[step:Transfer the conclusion back from the CW model to the original inclusion]Let \begin{align*} \Phi: (Z,Y) \longrightarrow (X,Y) \end{align*} be the homotopy equivalence of pairs under $Y$ supplied by the Lefschetz cell-attachment theorem. Thus, for the inclusion \begin{align*} j: Y \hookrightarrow Z, \end{align*} the composite $\Phi \circ j$ is homotopic to the original inclusion \begin{align*} i: Y \hookrightarrow X. \end{align*} Choose a base point $y_0 \in Y$. The map $\Phi: Z \to X$ is a homotopy equivalence, so the induced map \begin{align*} \Phi_*: \pi_k(Z,y_0) \longrightarrow \pi_k(X,y_0) \end{align*} is an isomorphism for every $k \geq 1$. Functoriality of homotopy groups and the homotopy $\Phi \circ j \simeq i$ give \begin{align*} \Phi_* \circ j_* = i_*: \pi_k(Y,y_0) \longrightarrow \pi_k(X,y_0), \end{align*} up to the standard basepoint-change identification determined by the homotopy. Hence $i_*$ is the composite of $j_*$ with an isomorphism. Therefore $i_*$ is an isomorphism for $1 \leq k < n-1$ and a surjection for $k = n-1$. The same argument applied to path components gives the asserted degree-$0$ statement, with $\pi_0$ interpreted as a pointed set of path components rather than a group. This completes the proof.[/step]

[guided]We now return from the CW model to the original inclusion. The compatibility required here is part of the Lefschetz cell-attachment input used at the beginning: it gives a homotopy equivalence of pairs under $Y$ \begin{align*} \Phi: (Z,Y) \longrightarrow (X,Y). \end{align*} If \begin{align*} j: Y \hookrightarrow Z \end{align*} is the CW-pair inclusion and \begin{align*} i: Y \hookrightarrow X \end{align*} is the original hyperplane-section inclusion, then "under $Y$" means that the composite $\Phi \circ j$ is homotopic to $i$ as a map $Y \to X$. Choose a base point $y_0 \in Y$. Since $\Phi: Z \to X$ is a homotopy equivalence, it induces an isomorphism \begin{align*} \Phi_*: \pi_k(Z,y_0) \longrightarrow \pi_k(X,y_0) \end{align*} for every $k \geq 1$. Homotopy groups are functorial: applying them to the composite $\Phi \circ j$ gives $\Phi_* \circ j_*$. Since $\Phi \circ j$ is homotopic to $i$, homotopy invariance gives \begin{align*} \Phi_* \circ j_* = i_*: \pi_k(Y,y_0) \longrightarrow \pi_k(X,y_0), \end{align*} up to the standard basepoint-change identification determined by the homotopy. This identity is exactly what is needed. For $1 \leq k < n-1$, the previous step proved that \begin{align*} j_*: \pi_k(Y,y_0) \longrightarrow \pi_k(Z,y_0) \end{align*} is an isomorphism. Composing an isomorphism with the isomorphism $\Phi_*$ gives an isomorphism, so $i_* = \Phi_* \circ j_*$ is an isomorphism. For $k = n-1$, the previous step proved that $j_*$ is surjective. Composing a surjection with the isomorphism $\Phi_*$ gives a surjection, so $i_*$ is surjective in degree $n-1$. For $k = 0$, the same functorial argument applies to path components, with $\pi_0$ interpreted as a pointed set rather than a group. This completes the transfer from the CW model to the original pair $(X,Y)$.[/guided]