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I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:

Give a proof of the nondegeneracy of the general hyperplane section of an projective irredicible nondegenerated variety $X$ of degree $\ge 2$ (...that's just to avoid the case $X$ linear; see also EDIT below) over $k= \mathbb{C}$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $X$ spans a $k$-plane, we have a rational map

$$ \varphi: (\mathbb{P}^{n})^* \dashrightarrow \mathbb{G}(k,n)$$

defined by sending a general hyperplane $H \in (\mathbb{P}^{n})^*$ to the span of $H \cap X$. Next, use the fact that the universal hyperplane section $\Omega_X$ (for definition see p 43 or below) of $X$ is irreducible to deduce that for any $H \in (\mathbb{P}^{n})^*$ and any point $\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$ (that is, any point in the image of the fiber of the graph $\Gamma_{\varphi}$, over $H$), the hyperplane section $H \cap X$ lies on the $k$-plane $\Lambda$. It follows that if the general hyperplane section of $X$ is degenerate, then all are - but any $n$ independent points of $X$ will span a hyperplane $H$ with $X \cap H $nondegenerate, contradiction.

Remark: The universal hyperplane section $\Omega_X \subset X \times (\mathbb{P}^{n})^*$ is defined as subvariety $\{(p, H) \ \vert \ p \in H \cap X \}$ und is irreducible (Theorem 5.8, page 53).

Question: I not understand how the irreducibility of $\Omega_X$ can be related to this problem, ie to use it here to conclude that $H \cap X \subset \Lambda$ for every $\Lambda \in \varphi(H)$.

Some thoughts: Let $U \subset (\mathbb{P}^{n})^*$ the open dense subset where $\varphi$ is regular. Since $\Gamma_{\varphi}$ is closure of the graph $\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$, it follows that $\Gamma_{\varphi}$ is irreducible and it suffice to find a closed subset of $\Gamma_{\varphi}$ which contains $\Gamma_U$.
A natural choice seems to take the closed subset $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ and intersect it with $\Gamma_{\varphi}$. By contruction it contains $\Gamma_U$.

But this approach nowhere makes use of the universal hyperplane section $\Omega_X$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.

EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $X$ is linear, let's do it and add the additional assumption that the degree of $X$ is $\ge 2$.

EDIT 2: actually I noticed that it could happen that that the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ might be not always closed inside $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $\Omega_X$?

My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $B \subset \Omega_X$ carrying informations about $\Gamma_{\varphi}$ and showing equality $B = \Omega_X$ would imply consequently the claim that $ H \cap X \subset \Lambda$ for all $ \Lambda \in \varphi(H)$. And the task becomes to show that this $B$ is closed...

Here is also the most naive candidate: set $B:= \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$. As I said, that's a very naive guess :) But it is of course dense in $\Omega_X$ and the big question becomes if this subset is closed?

What I can show that the set $ \{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$ is closed as image under proper projection map $p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^* $ of closed subset

$$ p_{12}^{-1}(\Omega_X) \cap p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi}) $$

but the conclusion that this set indeed equals $\Omega_X$ implies just that for every $H$ there exist some $\Lambda \in \varphi(H)$ with $ H \cap X \subset \Lambda$, but that's not enough.

I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:

Give a proof of the nondegeneracy of the general hyperplane section of an projective irreducible nondegenerated variety $X \subset \mathbb{P}^n$ of degree $\ge 2$ (...that's just to avoid the case $X$ linear; see also EDIT below) over $k= \mathbb{C}$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $X$ spans a $k$-plane, we have a rational map

$$ \varphi: (\mathbb{P}^{n})^* \dashrightarrow \mathbb{G}(k,n)$$

defined by sending a general hyperplane $H \in (\mathbb{P}^{n})^*$ to the span of $H \cap X$. Next, use the fact that the universal hyperplane section $\Omega_X$ (for definition see p 43 or below) of $X$ is irreducible to deduce that for any $H \in (\mathbb{P}^{n})^*$ and any point $\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$ (that is, any point in the image of the fiber of the graph $\Gamma_{\varphi}$, over $H$), the hyperplane section $H \cap X$ lies on the $k$-plane $\Lambda$. It follows that if the general hyperplane section of $X$ is degenerate, then all are - but any $n$ independent points of $X$ will span a hyperplane $H$ with $X \cap H $nondegenerate, contradiction.

Remark: The universal hyperplane section $\Omega_X \subset X \times (\mathbb{P}^{n})^*$ is defined as subvariety $\{(p, H) \ \vert \ p \in H \cap X \}$ und is irreducible (Theorem 5.8, page 53).

Question: I not understand how the irreducibility of $\Omega_X$ can be related to this problem, ie to use it here to conclude that $H \cap X \subset \Lambda$ for every $\Lambda \in \varphi(H)$.

Some thoughts: Let $U \subset (\mathbb{P}^{n})^*$ the open dense subset where $\varphi$ is regular. Since $\Gamma_{\varphi}$ is closure of the graph $\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$, it follows that $\Gamma_{\varphi}$ is irreducible and it suffice to find a closed subset of $\Gamma_{\varphi}$ which contains $\Gamma_U$.
A natural choice seems to take the closed subset $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ and intersect it with $\Gamma_{\varphi}$. By contruction it contains $\Gamma_U$.

But this approach nowhere makes use of the universal hyperplane section $\Omega_X$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.

EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $X$ is linear, let's do it and add the additional assumption that the degree of $X$ is $\ge 2$.

EDIT 2: actually I noticed that it could happen that that the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ might be not always closed inside $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $\Omega_X$?

My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $B \subset \Omega_X$ carrying informations about $\Gamma_{\varphi}$ and showing equality $B = \Omega_X$ would imply consequently the claim that $ H \cap X \subset \Lambda$ for all $ \Lambda \in \varphi(H)$. And the task becomes to show that this $B$ is closed...

Here is also the most naive candidate: set $B:= \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$. As I said, that's a very naive guess :) But it is of course dense in $\Omega_X$ and the big question becomes if this subset is closed?

What I can show that the set $ \{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$ is closed as image under proper projection map $p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^* $ of closed subset

$$ p_{12}^{-1}(\Omega_X) \cap p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi}) $$

but the conclusion that this set indeed equals $\Omega_X$ implies just that for every $H$ there exist some $\Lambda \in \varphi(H)$ with $ H \cap X \subset \Lambda$, but that's not enough.

I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:

Give a proof of the nondegeneracy of the general hyperplane section of an projective irredicible nondegenerated variety $X$ of degree $\ge 2$ (...that's just to avoid the case $X$ linear; see also EDIT below) over $k= \mathbb{C}$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $X$ spans a $k$-plane, we have a rational map

$$ \varphi: (\mathbb{P}^{n})^* \dashrightarrow \mathbb{G}(k,n)$$

defined by sending a general hyperplane $H \in (\mathbb{P}^{n})^*$ to the span of $H \cap X$. Next, use the fact that the universal hyperplane section $\Omega_X$ (for definition see p 43 or below) of $X$ is irreducible to deduce that for any $H \in (\mathbb{P}^{n})^*$ and any point $\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$ (that is, any point in the image of the fiber of the graph $\Gamma_{\varphi}$, over $H$), the hyperplane section $H \cap X$ lies on the $k$-plane $\Lambda$. It follows that if the general hyperplane section of $X$ is degenerate, then all are - but any $n$ independent points of $X$ will span a hyperplane $H$ with $X \cap H $nondegenerate, contradiction.

Remark: The universal hyperplane section $\Omega_X \subset X \times (\mathbb{P}^{n})^*$ is defined as subvariety $\{(p, H) \ \vert \ p \in H \cap X \}$ und is irreducible (Theorem 5.8, page 53).

Question: I not understand how the irreducibility of $\Omega_X$ can be related to this problem, ie to use it here to conclude that $H \cap X \subset \Lambda$ for every $\Lambda \in \varphi(H)$.

Some thoughts: Let $U \subset (\mathbb{P}^{n})^*$ the open dense subset where $\varphi$ is regular. Since $\Gamma_{\varphi}$ is closure of the graph $\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$, it follows that $\Gamma_{\varphi}$ is irreducible and it suffice to find a closed subset of $\Gamma_{\varphi}$ which contains $\Gamma_U$.
A natural choice seems to take the closed subset $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ and intersect it with $\Gamma_{\varphi}$. By contruction it contains $\Gamma_U$.

But this approach nowhere makes use of the universal hyperplane section $\Omega_X$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.

EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $X$ is linear, let's do it and add the additional assumption that the degree of $X$ is $\ge 2$.

EDIT 2: actually I noticed that it could happen that that the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ might be not always closed inside $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $\Omega_X$?

My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $B \subset \Omega_X$ carrying informations about $\Gamma_{\varphi}$ and showing equality $B = \Omega_X$ would imply consequently the claim that $ H \cap X \subset \Lambda$ for all $ \Lambda \in \varphi(H)$. And the task becomes to show that this $B$ is closed...

Here is also the most naive candidate for $B$: $ \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$. As I said, that's a naive guess :) But it is of course dense in $\Omega_X$ and the big question becomes if this subset is closed?

What I can show that the set $ \{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$ is closed as image under proper projection map $p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^* $ of closed subset

$$ p_{12}^{-1}(\Omega_X) \cap p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi}) $$

but the conclusion that this set indeed equals $\Omega_X$ implies just that for every $H$ there exist some $\Lambda \in \varphi(H)$ with $ H \cap X \subset \Lambda$, but that's not enough.

I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:

Give a proof of the nondegeneracy of the general hyperplane section of an projective irredicible nondegenerated variety $X$ of degree $\ge 2$ (...that's just to avoid the case $X$ linear; see also EDIT below) over $k= \mathbb{C}$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $X$ spans a $k$-plane, we have a rational map

$$ \varphi: (\mathbb{P}^{n})^* \dashrightarrow \mathbb{G}(k,n)$$

defined by sending a general hyperplane $H \in (\mathbb{P}^{n})^*$ to the span of $H \cap X$. Next, use the fact that the universal hyperplane section $\Omega_X$ (for definition see p 43 or below) of $X$ is irreducible to deduce that for any $H \in (\mathbb{P}^{n})^*$ and any point $\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$ (that is, any point in the image of the fiber of the graph $\Gamma_{\varphi}$, over $H$), the hyperplane section $H \cap X$ lies on the $k$-plane $\Lambda$. It follows that if the general hyperplane section of $X$ is degenerate, then all are - but any $n$ independent points of $X$ will span a hyperplane $H$ with $X \cap H $nondegenerate, contradiction.

Remark: The universal hyperplane section $\Omega_X \subset X \times (\mathbb{P}^{n})^*$ is defined as subvariety $\{(p, H) \ \vert \ p \in H \cap X \}$ und is irreducible (Theorem 5.8, page 53).

Question: I not understand how the irreducibility of $\Omega_X$ can be related to this problem, ie to use it here to conclude that $H \cap X \subset \Lambda$ for every $\Lambda \in \varphi(H)$.

Some thoughts: Let $U \subset (\mathbb{P}^{n})^*$ the open dense subset where $\varphi$ is regular. Since $\Gamma_{\varphi}$ is closure of the graph $\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$, it follows that $\Gamma_{\varphi}$ is irreducible and it suffice to find a closed subset of $\Gamma_{\varphi}$ which contains $\Gamma_U$.
A natural choice seems to take the closed subset $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ and intersect it with $\Gamma_{\varphi}$. By contruction it contains $\Gamma_U$.

But this approach nowhere makes use of the universal hyperplane section $\Omega_X$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.

EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $X$ is linear, let's do it and add the additional assumption that the degree of $X$ is $\ge 2$.

EDIT 2: actually I noticed that it could happen that that the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ might be not always closed inside $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $\Omega_X$?

My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $B \subset \Omega_X$ carrying informations about $\Gamma_{\varphi}$ and showing equality $B = \Omega_X$ would imply consequently the claim that $ H \cap X \subset \Lambda$ for all $ \Lambda \in \varphi(H)$. And the task becomes to show that this $B$ is closed...

Here is also the most naive candidate: set $B:= \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$. As I said, that's a very naive guess :) But it is of course dense in $\Omega_X$ and the big question becomes if this subset is closed?

What I can show that the set $ \{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$ is closed as image under proper projection map $p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^* $ of closed subset

$$ p_{12}^{-1}(\Omega_X) \cap p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi}) $$

but the conclusion that this set indeed equals $\Omega_X$ implies just that for every $H$ there exist some $\Lambda \in \varphi(H)$ with $ H \cap X \subset \Lambda$, but that's not enough.

I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:

Give a proof of the nondegeneracy of the general hyperplane section of an projective irredicible nondegenerated variety $X$ of degree $\ge 2$ (...that's just to avoid the case $X$ linear; see also EDIT below) over $k= \mathbb{C}$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $X$ spans a $k$-plane, we have a rational map

$$ \varphi: (\mathbb{P}^{n})^* \dashrightarrow \mathbb{G}(k,n)$$

defined by sending a general hyperplane $H \in (\mathbb{P}^{n})^*$ to the span of $H \cap X$. Next, use the fact that the universal hyperplane section $\Omega_X$ (for definition see p 43 or below) of $X$ is irreducible to deduce that for any $H \in (\mathbb{P}^{n})^*$ and any point $\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$ (that is, any point in the image of the fiber of the graph $\Gamma_{\varphi}$, over $H$), the hyperplane section $H \cap X$ lies on the $k$-plane $\Lambda$. It follows that if the general hyperplane section of $X$ is degenerate, then all are - but any $n$ independent points of $X$ will span a hyperplane $H$ with $X \cap H $nondegenerate, contradiction.

Remark: The universal hyperplane section $\Omega_X \subset X \times (\mathbb{P}^{n})^*$ is defined as subvariety $\{(p, H) \ \vert \ p \in H \cap X \}$ und is irreducible (Theorem 5.8, page 53).

Question: I not understand how the irreducibility of $\Omega_X$ can be related to this problem, ie to use it here to conclude that $H \cap X \subset \Lambda$ for every $\Lambda \in \varphi(H)$.

Some thoughts: Let $U \subset (\mathbb{P}^{n})^*$ the open dense subset where $\varphi$ is regular. Since $\Gamma_{\varphi}$ is closure of the graph $\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$, it follows that $\Gamma_{\varphi}$ is irreducible and it suffice to find a closed subset of $\Gamma_{\varphi}$ which contains $\Gamma_U$.
A natural choice seems to take the closed subset $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ and intersect it with $\Gamma_{\varphi}$. By contruction it contains $\Gamma_U$.

But this approach nowhere makes use of the universal hyperplane section $\Omega_X$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.

EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $X$ is linear, let's do it and add the additional assumption that the degree of $X$ is $\ge 2$.

EDIT 2: actually I noticed that it could happen that that the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ might be not always closed inside $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $\Omega_X$?

My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $B \subset \Omega_X$ carrying informations about $\Gamma_{\varphi}$ and showing equality $B = \Omega_X$ would imply consequently the claim that $ H \cap X \subset \Lambda$ for all $ \Lambda \in \varphi(H)$. And the task becomes to show that this this $B$ is closed...

Here is also the most naive candidate for $B$: $ \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$. As I said, that's a naive guess :) But it is of course dense in $\Omega_X$ and the big question becomes if this subset is closed?

What I can show that the set $ \{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$ is closed as image under proper projection map $p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^* $ of closed subset

$$ p_{12}^{-1}(\Omega_X) \cap p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi}) $$

but the conclusion that this set indeed equals $\Omega_X$ implies just that for every $H$ there exist some $\Lambda \in \varphi(H)$ with $ H \cap X \subset \Lambda$, but that's not enough.

I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:

Give a proof of the nondegeneracy of the general hyperplane section of an projective irredicible nondegenerated variety $X$ of degree $\ge 2$ (...that's just to avoid the case $X$ linear; see also EDIT below) over $k= \mathbb{C}$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $X$ spans a $k$-plane, we have a rational map

$$ \varphi: (\mathbb{P}^{n})^* \dashrightarrow \mathbb{G}(k,n)$$

defined by sending a general hyperplane $H \in (\mathbb{P}^{n})^*$ to the span of $H \cap X$. Next, use the fact that the universal hyperplane section $\Omega_X$ (for definition see p 43 or below) of $X$ is irreducible to deduce that for any $H \in (\mathbb{P}^{n})^*$ and any point $\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$ (that is, any point in the image of the fiber of the graph $\Gamma_{\varphi}$, over $H$), the hyperplane section $H \cap X$ lies on the $k$-plane $\Lambda$. It follows that if the general hyperplane section of $X$ is degenerate, then all are - but any $n$ independent points of $X$ will span a hyperplane $H$ with $X \cap H $nondegenerate, contradiction.

Remark: The universal hyperplane section $\Omega_X \subset X \times (\mathbb{P}^{n})^*$ is defined as subvariety $\{(p, H) \ \vert \ p \in H \cap X \}$ und is irreducible (Theorem 5.8, page 53).

Question: I not understand how the irreducibility of $\Omega_X$ can be related to this problem, ie to use it here to conclude that $H \cap X \subset \Lambda$ for every $\Lambda \in \varphi(H)$.

Some thoughts: Let $U \subset (\mathbb{P}^{n})^*$ the open dense subset where $\varphi$ is regular. Since $\Gamma_{\varphi}$ is closure of the graph $\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$, it follows that $\Gamma_{\varphi}$ is irreducible and it suffice to find a closed subset of $\Gamma_{\varphi}$ which contains $\Gamma_U$.
A natural choice seems to take the closed subset $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ and intersect it with $\Gamma_{\varphi}$. By contruction it contains $\Gamma_U$.

But this approach nowhere makes use of the universal hyperplane section $\Omega_X$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.

EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $X$ is linear, let's do it and add the additional assumption that the degree of $X$ is $\ge 2$.

EDIT 2: actually I noticed that it could happen that that the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ might be not always closed inside $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $\Omega_X$?

My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $B \subset \Omega_X$ carrying informations about $\Gamma_{\varphi}$ and showing equality $B = \Omega_X$ would imply consequently the claim that $ H \cap X \subset \Lambda$ for all $ \Lambda \in \varphi(H)$. And the task becomes to show that this $B$ is closed...

Here is also the most naive candidate for $B$: $ \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$. As I said, that's a naive guess :) But it is of course dense in $\Omega_X$ and the big question becomes if this subset is closed?

What I can show that the set $ \{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$ is closed as image under proper projection map $p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^* $ of closed subset

$$ p_{12}^{-1}(\Omega_X) \cap p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi}) $$

but the conclusion that this set indeed equals $\Omega_X$ implies just that for every $H$ there exist some $\Lambda \in \varphi(H)$ with $ H \cap X \subset \Lambda$, but that's not enough.