Return to Question

The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1]

The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (Proof: let $L_1$ and $L_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L_1]$ and $f[L_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L_1]$ and $f[L_2]$. This line intersects $f[L_1]$ in some point, say $f(p_1)$, and it intersects $f[L_2]$ in some point, say $f(p_2)$. The image under $f$ of the line through $p_1$ and $p_2$ must be the line through $f(p_1)$ and $f(p_2)$, which contains $p$, and therefore $p$ is in the image of $f$.)

My question is: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation?

Added later: Thanks to @Wojowu in the comments for explaining how to construct a function $\mathbb{R}^2\to\mathbb{R}$ that maps every line to the whole of $\mathbb{R}$, which answers the second question I asked.

So: what if we also insist that there be three non-collinear points in the image of $f$? It follows that there must be two distinct lines in the image of $f$, and then the argument above (more or less) shows that $f$ is surjective. But must it be injective?

[1] For a proof see e.g. Andrew Putman, The fundamental theorem of projective geometry.

The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1]

The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (Proof: let $L_1$ and $L_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L_1]$ and $f[L_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L_1]$ and $f[L_2]$. This line intersects $f[L_1]$ in some point, say $f(p_1)$, and it intersects $f[L_2]$ in some point, say $f(p_2)$. The image under $f$ of the line through $p_1$ and $p_2$ must be the line through $f(p_1)$ and $f(p_2)$, which contains $p$, and therefore $p$ is in the image of $f$.)

My question is: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation?

Added later: Thanks to @Wojowu in the comments for explaining how to construct a function $\mathbb{R}^2\to\mathbb{R}$ that maps every line to the whole of $\mathbb{R}$, which answers the second question I asked.

So: what if we also insist that there be three non-collinear points in the image of $f$? It follows that there must be two distinct lines in the image of $f$, and then the argument above (more or less) shows that $f$ is surjective. But must it be injective?

[1] For a proof of the general case see e.g. Andrew Putman, The fundamental theorem of projective geometry.

The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1]

The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (Proof: let $L_1$ and $L_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L_1]$ and $f[L_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L_1]$ and $f[L_2]$. This line intersects $f[L_1]$ in some point, say $f(p_1)$, and it intersects $f[L_2]$ in some point, say $f(p_2)$. The image under $f$ of the line through $p_1$ and $p_2$ must be the line through $f(p_1)$ and $f(p_2)$, which contains $p$, and therefore $p$ is in the image of $f$.)

My question is: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation?

Edit: Thanks to @Wojowu in the comments for explaining how to construct a function $\mathbb{R}^2\to\mathbb{R}$ that maps every line to the whole of $\mathbb{R}$, which answers the second question I asked.

So: what if we also insist that there be three non-collinear points in the image of $f$?

[1] For a proof see e.g. Andrew Putman, The fundamental theorem of projective geometry.

The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1]

The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (Proof: let $L_1$ and $L_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L_1]$ and $f[L_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L_1]$ and $f[L_2]$. This line intersects $f[L_1]$ in some point, say $f(p_1)$, and it intersects $f[L_2]$ in some point, say $f(p_2)$. The image under $f$ of the line through $p_1$ and $p_2$ must be the line through $f(p_1)$ and $f(p_2)$, which contains $p$, and therefore $p$ is in the image of $f$.)

My question is: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation?

Added later: Thanks to @Wojowu in the comments for explaining how to construct a function $\mathbb{R}^2\to\mathbb{R}$ that maps every line to the whole of $\mathbb{R}$, which answers the second question I asked.

So: what if we also insist that there be three non-collinear points in the image of $f$? It follows that there must be two distinct lines in the image of $f$, and then the argument above (more or less) shows that $f$ is surjective. But must it be injective?

[1] For a proof see e.g. Andrew Putman, The fundamental theorem of projective geometry.

The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1]

The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (Proof: let $L_1$ and $L_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L_1]$ and $f[L_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L_1]$ and $f[L_2]$. This line intersects $f[L_1]$ in some point, say $f(p_1)$, and it intersects $f[L_2]$ in some point, say $f(p_2)$. The image under $f$ of the line through $p_1$ and $p_2$ must be the line through $f(p_1)$ and $f(p_2)$, which contains $p$, and therefore $p$ is in the image of $f$.)

My question is: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation?

[1] For a proof see e.g. Andrew Putman, The fundamental theorem of projective geometry.

The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1]

The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (Proof: let $L_1$ and $L_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L_1]$ and $f[L_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L_1]$ and $f[L_2]$. This line intersects $f[L_1]$ in some point, say $f(p_1)$, and it intersects $f[L_2]$ in some point, say $f(p_2)$. The image under $f$ of the line through $p_1$ and $p_2$ must be the line through $f(p_1)$ and $f(p_2)$, which contains $p$, and therefore $p$ is in the image of $f$.)

My question is: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation?

Edit: Thanks to @Wojowu in the comments for explaining how to construct a function $\mathbb{R}^2\to\mathbb{R}$ that maps every line to the whole of $\mathbb{R}$, which answers the second question I asked.

So: what if we also insist that there be three non-collinear points in the image of $f$?

[1] For a proof see e.g. Andrew Putman, The fundamental theorem of projective geometry.