There is a common proof strategy in algebraic geometry which works like this. Here is an example:

Problem: Show that the variety of smooth plane cubics which have flexes at $(0:0:1)$, $(0:1:0)$, $(1:0:0)$ and $(1:1:1)$ is smooth.

Now, you may or may not be able to do this by direct computation, but I think you will agree that is nontrivial. So the following slick proof may impress you.

Proof: Let $U$ be the variety of smooth cubics (an open subset of $\mathbb{CP}^9$). Let $X \subset U \times (\mathbb{CP}^2)^4$ be the variety of ordered pairs: (smooth cubic, quadruple of flexes). We are interested in the fiber of $X$ over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. I claim that the easiest way to show that this is smooth is to use the whole family $X$.

Since a smooth cubic has $9$ distinct flexes, $X$ is a $9^4$-fold cover of $U$, and is thus smooth. Let $\pi$ be the projection $X \to (\mathbb{CP}^2)^4$. By Sard's theorem (or the analogous algebraic result), the generic fiber of $\pi$ is smooth.

But the group $PGL_3$ acts compatibly on $X$ and $(\mathbb{CP}^2)^4$, so the set of points $z$ in $(\mathbb{CP}^2)^4$ for which $\pi^{-1}(z)$ is smooth is $PGL_3$-invariant. The $PGL_3$ orbit through $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is open and dense, so any dense $PGL_3$ invariant set must meet $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. So we deduce that the fiber over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is smooth, as desired.

There is a common proof strategy in algebraic geometry which works like this. Here is an example:

Problem: Show that the variety of smooth plane cubics which have flexes at $(0:0:1)$, $(0:1:0)$, $(1:0:0)$ and $(1:1:1)$ is smooth.

Now, you may or may not be able to do this by direct computation, but I think you will agree that is nontrivial. So the following slick proof may impress you.

Proof: Let $U$ be the variety of smooth cubics (an open subset of $\mathbb{CP}^9$). Let $X \subset U \times (\mathbb{CP}^2)^4$ be the variety of ordered pairs: (smooth cubic, quadruple of flexes). We are interested in the fiber of $X$ over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. I claim that the easiest way to show that this is smooth is to use the whole family $X$.

Since a smooth cubic has $9$ distinct flexes, $X$ is a $9^4$-fold cover of $U$, and is thus smooth. Let $\pi$ be the projection $X \to (\mathbb{CP}^2)^4$. By Sard's theorem (or the analogous algebraic result), the generic fiber of $\pi$ is smooth.

But the group $PGL_3$ acts compatibly on $X$ and $(\mathbb{CP}^2)^4$, so the set of points $z$ in $(\mathbb{CP}^2)^4$ for which $\pi^{-1}(z)$ is smooth is $PGL_3$-invariant. The $PGL_3$ orbit through $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is open and dense, so any dense $PGL_3$ invariant set must meet $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. So we deduce that the fiber over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is smooth, as desired.

There is a common proof strategy in algebraic geometry which works like this. Here is an example:

Problem: Show that the variety of smooth plane cubics which have flexes at $(0:0:1)$, $(0:1:0)$, $(1:0:0)$ and $(1:1:1)$ is smooth.

Now, you may or may not be able to o this by direct computation, but I think you will agree that is nontrivial. So the following slick proof may impress you.

Proof: Let $U$ be the variety of smooth cubics (an open subset of $\mathbb{CP}^9$). Let $X \subset U \times (\mathbb{CP}^2)^4$ be the variety of ordered pairs: (smooth cubic, quadruple of flexes). We are interested in the fiber of $X$ over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. I claim that the easiest way to show that this is smooth is to use the whole family $X$.

Since a smooth cubic has $9$ distinct flexes, $X$ is a $9^4$-fold cover of $U$, and is thus smooth. Let $\pi$ be the projection $X \to (\mathbb{CP}^2)^4$. By Sard's theorem (or the analogous algebraic result), the generic fiber of $\pi$ is smooth.

But the group $PGL_3$ acts compatibly on $X$ and $(\mathbb{CP}^2)^4$, so the set of points $z$ in $(\mathbb{CP}^2)^4$ for which $\pi^{-1}(z)$ is smooth is $PGL_3$-invariant. The $PGL_3$ orbit through $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is open and dense, so any dense $PGL_3$ invariant set must meet $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. So we deduce that the fiber over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is smooth, as desired.

There is a common proof strategy in algebraic geometry which works like this. Here is an example:

Problem: Show that the variety of smooth plane cubics which have flexes at $(0:0:1)$, $(0:1:0)$, $(1:0:0)$ and $(1:1:1)$ is smooth.

Now, you may or may not be able to do this by direct computation, but I think you will agree that is nontrivial. So the following slick proof may impress you.

Proof: Let $U$ be the variety of smooth cubics (an open subset of $\mathbb{CP}^9$). Let $X \subset U \times (\mathbb{CP}^2)^4$ be the variety of ordered pairs: (smooth cubic, quadruple of flexes). We are interested in the fiber of $X$ over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. I claim that the easiest way to show that this is smooth is to use the whole family $X$.

Since a smooth cubic has $9$ distinct flexes, $X$ is a $9^4$-fold cover of $U$, and is thus smooth. Let $\pi$ be the projection $X \to (\mathbb{CP}^2)^4$. By Sard's theorem (or the analogous algebraic result), the generic fiber of $\pi$ is smooth.

But the group $PGL_3$ acts compatibly on $X$ and $(\mathbb{CP}^2)^4$, so the set of points $z$ in $(\mathbb{CP}^2)^4$ for which $\pi^{-1}(z)$ is smooth is $PGL_3$-invariant. The $PGL_3$ orbit through $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is open and dense, so any dense $PGL_3$ invariant set must meet $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. So we deduce that the fiber over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is smooth, as desired.