Dear anonymous, let $ X \subset \mathbb P^n(\mathbb C)$ be a smooth hypersurface of degree $d$ in projective space. Let me show how Lefschetz directly calculates the homology $H_i(X)=H_i(X,\mathbb Z)$ of $X$, except for $H_{n-1}(X)$.

There is an embedding $v:\mathbb P^n(\mathbb C) \to \mathbb P^N(\mathbb C)$, called the Veronese embedding, mapping the original $P^n(\mathbb C)$ to a huge projective space $\mathbb P^N(\mathbb C)$ of dimension $N=\binom{n+d}{d} -1$
The image of $v$ is then an isomorphic copy $P$ of $\mathbb P^n(\mathbb C)$ and the charm of Veronese's embedding is that $v(X)$ is now a hyperplane section of $P $ i.e. $v(X)=P \cap H$ for some hyperplane $H \subset \mathbb P ^N(\mathbb C)$ .

You thus get $H_i(X) \simeq H_i(P)$ for $i\leq n-2$ by applying Lefschetz for homology. For $i\geq n$ you deduce (by applying Poincaré duality and Lefschetz for cohomology ) $H_i(X) \simeq H^{2n-2-i}(X)\simeq H^{2n-2-i}(\mathbb P^n(\mathbb C))$. Since $P\simeq P^n(\mathbb C)$ all the above homology groups are easy to calculate. Indeed, for $0 \leq i \leq 2n$, projective space has homology $H_i(P^n(\mathbb C))$ alternately $0$ and $\mathbb Z$. Ditto for cohomology.

The only group which has partly escaped us is $H_{n-1}(X)$, for which Lefschetz only tells us that it surjects onto $H_{n-1}(P^n(\mathbb C))$. As a reality check, notice that for $n=2$ we know that the rank of $H_{1}(X)$ is $2g=(d-1)(d-2)$, where $g$ is the genus of our curve $X$ . This is indeed not equal to the rank of any (co)homology group of $P^2(\mathbb C)$ for $d \geq 3$

Optional complement: Allow me to show how concrete the Veronese embedding is. It maps the point $(x_0:x_1:x_2:\cdots:x_n)\in \mathbb P^n$ to the point $(\cdots:M_\delta (x):\cdots) \in \mathbb P^N$ where $\delta = (d_0,\cdots,d_n) \quad d_0+\cdots d_n=d$ and $M_\delta (x)=x_0^{d_0}\cdots x_n^{d_n}$.
So that if for example $n=d=2$, the conic $x^2+2y^2+3y^2-4xy-5xz-6yz=0$ in $\mathbb P^2$ gets mapped isomorphically by Veronese onto the hyperplane section of $v(\mathbb P^2)$ by the hyperplane $X+2Y+3Z-4U-5V-6W=0$ of $\mathbb P^5$ . What could be simpler?

Dear anonymous, let $ X \subset \mathbb P^n(\mathbb C)$ be a smooth hypersurface of degree $d$ in projective space. Let me show how Lefschetz directly calculates the homology $H_i(X)=H_i(X,\mathbb Z)$ of $X$, except for $H_{n-1}(X)$.

There is an embedding $v:\mathbb P^n(\mathbb C) \to \mathbb P^N(\mathbb C)$, called the Veronese embedding, mapping the original $P^n(\mathbb C)$ to a huge projective space $\mathbb P^N(\mathbb C)$ of dimension $N=\binom{n+d}{d} -1$
The image of $v$ is then an isomorphic copy $P$ of $\mathbb P^n(\mathbb C)$ and the charm of Veronese's embedding is that $v(X)$ is now a hyperplane section of $P $ i.e. $v(X)=P \cap H$ for some hyperplane $H \subset \mathbb P ^N(\mathbb C)$ .

You thus get $H_i(X) \simeq H_i(P)$ for $i\leq n-2$ by applying Lefschetz for homology. For $i\geq n$ you deduce (by applying Poincaré duality and Lefschetz for cohomology ) $H_i(X) \simeq H^{2n-2-i}(X)\simeq H^{2n-2-i}(\mathbb P^n(\mathbb C))$. Since $P\simeq P^n(\mathbb C)$ all the above homology groups are easy to calculate. Indeed, for $0 \leq i \leq 2n$, projective space has homology $H_i(P^n(\mathbb C))$ alternately $0$ and $\mathbb Z$. Ditto for cohomology.

The only group which has partly escaped us is $H_{n-1}(X)$, for which Lefschetz only tells us that it surjects onto $H_{n-1}(P^n(\mathbb C))$. As a reality check, notice that for $n=2$ we know that the rank of $H_{1}(X)$ is $2g=(d-1)(d-2)$, where $g$ is the genus of our curve $X$ . This is indeed not equal to the rank of any (co)homology group of $P^2(\mathbb C)$ for $d \geq 3$

Optional complement: Allow me to show how concrete the Veronese embedding is. It maps the point $(x_0:x_1:x_2:\cdots:x_n)\in \mathbb P^n$ to the point $(\cdots:M_\delta (x):\cdots) \in \mathbb P^N$ where $\delta = (d_0,\cdots,d_n) \quad d_0+\cdots d_n=d$ and $M_\delta (x)=x_0^{d_0}\cdots x_n^{d_n}$.
So that if for example $n=d=2$, the conic $x^2+2y^2+3y^2-4xy-5xz-6yz=0$ in $\mathbb P^2$ gets mapped isomorphically by Veronese onto the hyperplane section of the Veronese surface $v(\mathbb P^2)$ by the hyperplane $X+2Y+3Z-4U-5V-6W=0$ of $\mathbb P^5$ . What could be simpler?

Dear anonymous, let $ X \subset \mathbb P^n(\mathbb C)$ be a smooth hypersurface of degree $d$ in projective space. Let me show how Lefschetz directly calculates the homology $H_i(X)=H_i(X,\mathbb Z)$ of $X$, except for $H_{n-1}(X)$.

There is an embedding $v:\mathbb P^n(\mathbb C) \to \mathbb P^N(\mathbb C)$, called the Veronese embedding, mapping the original $P^n(\mathbb C)$ to a huge projective space $\mathbb P^N(\mathbb C)$ of dimension $N=\binom{n+d}{d} -1$
The image of $v$ is then an isomorphic copy $P$ of $\mathbb P^n(\mathbb C)$ and the charm of Veronese's embedding is that $v(X)$ is now a hyperplane section of $P $ i.e. $v(X)=P \cap H$ for some hyperplane $H \subset \mathbb P ^N(\mathbb C)$ .

You thus get $H_i(X) \simeq H_i(P)$ for $i\leq n-2$ by applying Lefschetz for homology. For $i\geq n$ you deduce (by applying Poincaré duality and Lefschetz for cohomology ) $H_i(X) \simeq H^{2n-2-i}(X)\simeq H^{2n-2-i}(\mathbb P^n(\mathbb C))$. Since $P\simeq P^n(\mathbb C)$ all the above homology groups are easy to calculate. Indeed, for $0 \leq i \leq 2n$, projective space has homology $H_i(P^n(\mathbb C))$ alternately $0$ and $\mathbb Z$. Ditto for cohomology.

The only group which has partly escaped us is $H_{n-1}(X)$, for which Lefschetz only tells us that it surjects onto $H_{n-1}(P^n(\mathbb C))$. As a reality check, notice that for $n=2$ we know that the rank of $H_{1}(X)$ is $2g=(d-1)(d-2)$, where $g$ is the genus of our curve $X$ . This is indeed not equal to the rank of any (co)homology group of $P^2(\mathbb C)$ for $d \geq 3$

Dear anonymous, let $ X \subset \mathbb P^n(\mathbb C)$ be a smooth hypersurface of degree $d$ in projective space. Let me show how Lefschetz directly calculates the homology $H_i(X)=H_i(X,\mathbb Z)$ of $X$, except for $H_{n-1}(X)$.

There is an embedding $v:\mathbb P^n(\mathbb C) \to \mathbb P^N(\mathbb C)$, called the Veronese embedding, mapping the original $P^n(\mathbb C)$ to a huge projective space $\mathbb P^N(\mathbb C)$ of dimension $N=\binom{n+d}{d} -1$
The image of $v$ is then an isomorphic copy $P$ of $\mathbb P^n(\mathbb C)$ and the charm of Veronese's embedding is that $v(X)$ is now a hyperplane section of $P $ i.e. $v(X)=P \cap H$ for some hyperplane $H \subset \mathbb P ^N(\mathbb C)$ .

You thus get $H_i(X) \simeq H_i(P)$ for $i\leq n-2$ by applying Lefschetz for homology. For $i\geq n$ you deduce (by applying Poincaré duality and Lefschetz for cohomology ) $H_i(X) \simeq H^{2n-2-i}(X)\simeq H^{2n-2-i}(\mathbb P^n(\mathbb C))$. Since $P\simeq P^n(\mathbb C)$ all the above homology groups are easy to calculate. Indeed, for $0 \leq i \leq 2n$, projective space has homology $H_i(P^n(\mathbb C))$ alternately $0$ and $\mathbb Z$. Ditto for cohomology.

The only group which has partly escaped us is $H_{n-1}(X)$, for which Lefschetz only tells us that it surjects onto $H_{n-1}(P^n(\mathbb C))$. As a reality check, notice that for $n=2$ we know that the rank of $H_{1}(X)$ is $2g=(d-1)(d-2)$, where $g$ is the genus of our curve $X$ . This is indeed not equal to the rank of any (co)homology group of $P^2(\mathbb C)$ for $d \geq 3$

Optional complement: Allow me to show how concrete the Veronese embedding is. It maps the point $(x_0:x_1:x_2:\cdots:x_n)\in \mathbb P^n$ to the point $(\cdots:M_\delta (x):\cdots) \in \mathbb P^N$ where $\delta = (d_0,\cdots,d_n) \quad d_0+\cdots d_n=d$ and $M_\delta (x)=x_0^{d_0}\cdots x_n^{d_n}$.
So that if for example $n=d=2$, the conic $x^2+2y^2+3y^2-4xy-5xz-6yz=0$ in $\mathbb P^2$ gets mapped isomorphically by Veronese onto the hyperplane section of $v(\mathbb P^2)$ by the hyperplane $X+2Y+3Z-4U-5V-6W=0$ of $\mathbb P^5$ . What could be simpler?