Show that the set of Endormorphism of a given graph is equal to the set of Automorphism.
Let $A$ be the $10 times 10 = [a_{ij}]$, where $a_{ij} = (i, j)$. Let $G$ be a graph whose vertex set is the eateries of the matrix $A$ and the two distinct vertex $(i, j)$ and $(k, l)$ are adjacent if
$$k neq j ; text{and} ; l neq i.$$
I have found lots of graph theoretic properties of this graph such that neighbourhood of every vertex, chromatic number clique number, independence number etc.
I want to prove that End$(G) = $Aut$(G)$.
I guess that End$(G) = $Aut$(G)$. We know that End$(G) subseteq$S-End$(G) subseteq$ Aut$(G)$, where S-End$(G)$ is the set of strong endomorphisn on a graph $G$. A endomorphism is said to be strong endomorphism if $f(x)$ is adjacent to $f(y)$ implies $x$ is adjacent to $y$. I found that S-End$(G) = $Aut$(G)$. We need to show that End$(G) = $S-End$(G)$.
I have also tried to prove my result by using the concept of core. A graph $G$ is said to be a core if and only if Aut$G$ = End$(G)$. If $H$ is a proper subgraph of $G$, then a homomorphism $f : G rightarrow H$ is called retraction if $f(h) = h ; forall h in H$ and $H$ is called retract. We know that a graph is a core if and only it admits no retract.
graph-theory algebraic-graph-theory
asked Dec 27 '18 at 4:25
Struggler
1,133718
add a comment |
Let $A$ be the $10 times 10 = [a_{ij}]$, where $a_{ij} = (i, j)$. Let $G$ be a graph whose vertex set is the eateries of the matrix $A$ and the two distinct vertex $(i, j)$ and $(k, l)$ are adjacent if
$$k neq j ; text{and} ; l neq i.$$
I have found lots of graph theoretic properties of this graph such that neighbourhood of every vertex, chromatic number clique number, independence number etc.
I want to prove that End$(G) = $Aut$(G)$.
I guess that End$(G) = $Aut$(G)$. We know that End$(G) subseteq$S-End$(G) subseteq$ Aut$(G)$, where S-End$(G)$ is the set of strong endomorphisn on a graph $G$. A endomorphism is said to be strong endomorphism if $f(x)$ is adjacent to $f(y)$ implies $x$ is adjacent to $y$. I found that S-End$(G) = $Aut$(G)$. We need to show that End$(G) = $S-End$(G)$.
I have also tried to prove my result by using the concept of core. A graph $G$ is said to be a core if and only if Aut$G$ = End$(G)$. If $H$ is a proper subgraph of $G$, then a homomorphism $f : G rightarrow H$ is called retraction if $f(h) = h ; forall h in H$ and $H$ is called retract. We know that a graph is a core if and only it admits no retract.
graph-theory algebraic-graph-theory
asked Dec 27 '18 at 4:25
Struggler
1,133718
Isn't $(i,j)mapsto (i,i)$ a weak endomorphism?
– Hagen von Eitzen
Dec 27 '18 at 5:12
This is not an endomorhism. Because $(1, 2)$ is adjacent with $(1, 3)$ but $f(1, 2) = (1, 1) = f(1, 3)$.
– Struggler
Dec 27 '18 at 6:12
add a comment |
Let $A$ be the $10 times 10 = [a_{ij}]$, where $a_{ij} = (i, j)$. Let $G$ be a graph whose vertex set is the eateries of the matrix $A$ and the two distinct vertex $(i, j)$ and $(k, l)$ are adjacent if
$$k neq j ; text{and} ; l neq i.$$
I have found lots of graph theoretic properties of this graph such that neighbourhood of every vertex, chromatic number clique number, independence number etc.
I want to prove that End$(G) = $Aut$(G)$.
I guess that End$(G) = $Aut$(G)$. We know that End$(G) subseteq$S-End$(G) subseteq$ Aut$(G)$, where S-End$(G)$ is the set of strong endomorphisn on a graph $G$. A endomorphism is said to be strong endomorphism if $f(x)$ is adjacent to $f(y)$ implies $x$ is adjacent to $y$. I found that S-End$(G) = $Aut$(G)$. We need to show that End$(G) = $S-End$(G)$.
I have also tried to prove my result by using the concept of core. A graph $G$ is said to be a core if and only if Aut$G$ = End$(G)$. If $H$ is a proper subgraph of $G$, then a homomorphism $f : G rightarrow H$ is called retraction if $f(h) = h ; forall h in H$ and $H$ is called retract. We know that a graph is a core if and only it admits no retract.
graph-theory algebraic-graph-theory
asked Dec 27 '18 at 4:25
Struggler
1,133718
Let $A$ be the $10 times 10 = [a_{ij}]$, where $a_{ij} = (i, j)$. Let $G$ be a graph whose vertex set is the eateries of the matrix $A$ and the two distinct vertex $(i, j)$ and $(k, l)$ are adjacent if
$$k neq j ; text{and} ; l neq i.$$
I have found lots of graph theoretic properties of this graph such that neighbourhood of every vertex, chromatic number clique number, independence number etc.
I want to prove that End$(G) = $Aut$(G)$.
I guess that End$(G) = $Aut$(G)$. We know that End$(G) subseteq$S-End$(G) subseteq$ Aut$(G)$, where S-End$(G)$ is the set of strong endomorphisn on a graph $G$. A endomorphism is said to be strong endomorphism if $f(x)$ is adjacent to $f(y)$ implies $x$ is adjacent to $y$. I found that S-End$(G) = $Aut$(G)$. We need to show that End$(G) = $S-End$(G)$.
I have also tried to prove my result by using the concept of core. A graph $G$ is said to be a core if and only if Aut$G$ = End$(G)$. If $H$ is a proper subgraph of $G$, then a homomorphism $f : G rightarrow H$ is called retraction if $f(h) = h ; forall h in H$ and $H$ is called retract. We know that a graph is a core if and only it admits no retract.
graph-theory algebraic-graph-theory
graph-theory algebraic-graph-theory
asked Dec 27 '18 at 4:25
Struggler
1,133718
asked Dec 27 '18 at 4:25
Struggler
1,133718
asked Dec 27 '18 at 4:25
Struggler
1,133718
asked Dec 27 '18 at 4:25
Struggler
1,133718
asked Dec 27 '18 at 4:25
Struggler
1,133718
1,133718
Isn't $(i,j)mapsto (i,i)$ a weak endomorphism?
– Hagen von Eitzen
Dec 27 '18 at 5:12
This is not an endomorhism. Because $(1, 2)$ is adjacent with $(1, 3)$ but $f(1, 2) = (1, 1) = f(1, 3)$.
– Struggler
Dec 27 '18 at 6:12
add a comment |
Isn't $(i,j)mapsto (i,i)$ a weak endomorphism?
– Hagen von Eitzen
Dec 27 '18 at 5:12
This is not an endomorhism. Because $(1, 2)$ is adjacent with $(1, 3)$ but $f(1, 2) = (1, 1) = f(1, 3)$.
– Struggler
Dec 27 '18 at 6:12
Isn't $(i,j)mapsto (i,i)$ a weak endomorphism?
– Hagen von Eitzen
Dec 27 '18 at 5:12
Isn't $(i,j)mapsto (i,i)$ a weak endomorphism?
– Hagen von Eitzen
Dec 27 '18 at 5:12
This is not an endomorhism. Because $(1, 2)$ is adjacent with $(1, 3)$ but $f(1, 2) = (1, 1) = f(1, 3)$.
– Struggler
Dec 27 '18 at 6:12
This is not an endomorhism. Because $(1, 2)$ is adjacent with $(1, 3)$ but $f(1, 2) = (1, 1) = f(1, 3)$.
– Struggler
Dec 27 '18 at 6:12
add a comment |
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Isn't $(i,j)mapsto (i,i)$ a weak endomorphism?
– Hagen von Eitzen
Dec 27 '18 at 5:12
This is not an endomorhism. Because $(1, 2)$ is adjacent with $(1, 3)$ but $f(1, 2) = (1, 1) = f(1, 3)$.
– Struggler
Dec 27 '18 at 6:12