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Alex Meiburg
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Given a graph $G$, and a bijective endomorphism $f$ (that is, a graph homeomorphism $f : G \to G$ that establishes a bijection on the vertices), it is true that $f$ is an automorphism whenever $|G|$ is finite.

When $G$ is no longer finite, this no longer holds: consider the graph whose vertices are the integers, with an edge between every neighboring pair of positive integers, and the endomorphism $f(n)=1+n$. The vertices are in bijection, all pairs of neighboring positive integers are stillget mapped to new pairs of neighboring positive integers, butso it's a homeomorphism. But the previously non-adjacent $(0,1) \to (f(0),f(1)) = (1,2)$ which is adjacent. So $f$ is not an isomorphism.

However, $f$ restricts to an isomorphism on most of its domain: restricting it to $H=\mathbb{Z}\setminus\{0\}$ or $\mathbb{Z}\setminus\{1\}$ produces an isomorphism. (Not an automorphism of subgraphs, because the domain and range are not the same.) This is a "large" restriction in the sense that $|G|=|H|$.

Given how nicely this works for finite graphs and all the infinite examples I could think of... can we always find such a restriction?

Given a graph $G$, and a bijective endomorphism $f$ (that is, a graph homeomorphism $f : G \to G$ that establishes a bijection on the vertices), it is true that $f$ is an automorphism whenever $|G|$ is finite.

When $G$ is no longer finite, this no longer holds: consider the graph whose vertices are the integers, with an edge between every neighboring pair of positive integers, and the endomorphism $f(n)=1+n$. The vertices are in bijection, all pairs of neighboring positive integers are still neighboring positive integers, but the previously non-adjacent $(0,1) \to (f(0),f(1)) = (1,2)$ which is adjacent. So $f$ is not an isomorphism.

However, $f$ restricts to an isomorphism on most of its domain: restricting it to $H=\mathbb{Z}\setminus\{0\}$ or $\mathbb{Z}\setminus\{1\}$ produces an isomorphism. (Not an automorphism of subgraphs, because the domain and range are not the same.) This is a "large" restriction in the sense that $|G|=|H|$.

Given how nicely this works for finite graphs and all the infinite examples I could think of... can we always find such a restriction?

Given a graph $G$, and a bijective endomorphism $f$ (that is, a graph homeomorphism $f : G \to G$ that establishes a bijection on the vertices), it is true that $f$ is an automorphism whenever $|G|$ is finite.

When $G$ is no longer finite, this no longer holds: consider the graph whose vertices are the integers, with an edge between every neighboring pair of positive integers, and the endomorphism $f(n)=1+n$. The vertices are in bijection, all pairs of neighboring positive integers get mapped to new pairs of neighboring positive integers, so it's a homeomorphism. But the previously non-adjacent $(0,1) \to (f(0),f(1)) = (1,2)$ which is adjacent. So $f$ is not an isomorphism.

However, $f$ restricts to an isomorphism on most of its domain: restricting it to $H=\mathbb{Z}\setminus\{0\}$ or $\mathbb{Z}\setminus\{1\}$ produces an isomorphism. (Not an automorphism of subgraphs, because the domain and range are not the same.) This is a "large" restriction in the sense that $|G|=|H|$.

Given how nicely this works for finite graphs and all the infinite examples I could think of... can we always find such a restriction?

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Alex Meiburg
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Does every bijective graph endomorphism restrict to a full-cardinality isomorphism?

Given a graph $G$, and a bijective endomorphism $f$ (that is, a graph homeomorphism $f : G \to G$ that establishes a bijection on the vertices), it is true that $f$ is an automorphism whenever $|G|$ is finite.

When $G$ is no longer finite, this no longer holds: consider the graph whose vertices are the integers, with an edge between every neighboring pair of positive integers, and the endomorphism $f(n)=1+n$. The vertices are in bijection, all pairs of neighboring positive integers are still neighboring positive integers, but the previously non-adjacent $(0,1) \to (f(0),f(1)) = (1,2)$ which is adjacent. So $f$ is not an isomorphism.

However, $f$ restricts to an isomorphism on most of its domain: restricting it to $H=\mathbb{Z}\setminus\{0\}$ or $\mathbb{Z}\setminus\{1\}$ produces an isomorphism. (Not an automorphism of subgraphs, because the domain and range are not the same.) This is a "large" restriction in the sense that $|G|=|H|$.

Given how nicely this works for finite graphs and all the infinite examples I could think of... can we always find such a restriction?