equivalence between the minor and topological minor of K_5 or K_{3,3}

equivalence between the minor and topological minor of

K

5

or

K

3

,

3

⇐

)

Remark that if a graph G contains

K

5

or

K

3

,

3

as a

topological minor

(

http://planetmath.org/subdivision

) it is easy to see that it also contains respectively

K

5

or

K

3

,

3

as a minor.

⇒

)

Same goes for

K

3

,

3

as minor. If graph G contains a minor of

K

3

,

3

as minor it also contains a topological minor (

http://planetmath.org/subdivision

) of

K

3

,

3

in G.
So it’s suffices to show that if graph G contains a

K

5

as minor, it also contains a

K

5

or

K

3

,

3

as topological minor (

http://planetmath.org/subdivision

).

Suppose that

G

⪰

K

5

, let

K

⊆

G

be minimal such that

K

is a minor of

G

. Because

K

is minimal every branch set of

K

induces a tree in

K

and every two branch sets have exactly one edge between them.
For every branch tree

V

x

we can add all four edges joining it to the other branches, this way we obtain a tree

T

x

.
If each of the trees

T

x

is a topological minor (

http://planetmath.org/subdivision

) of

K

1

,

4

,

K

will be a topological minor (

http://planetmath.org/subdivision

) of

K

5

. This image should make clear how our

K

looks like. The big circles are presenting different sets of vertices.

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⁢

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.

[

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⁢

r

⁢

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⁢

r

]

⁢

&

⁢

&

⁢

&

⁢

&

○

\ar

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.

[

l

⁢

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]

⁢

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.

[

u

⁢

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]

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\ar

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.

[

u

⁢

l

⁢

l

⁢

l

]

∙

\ar

⁢

@

-

[

u

]

⁢

&

∙

\ar

⁢

@

-

[

u

⁢

u

]

⁢

&

⁢

&

∙

\ar

⁢

@

-

[

u

⁢

u

]

⁢

&

∙

\ar

⁢

@

-

[

u

]

⁢

&

⁢

T

x

=

M

⁢

K

1

,

4

⁢

&

⁢

&

○

\ar

⁢

@

-

[

u

⁢

r

⁢

r

]

⁢

\ar

⁢

@

-

[

u

⁢

r

]

⁢

\ar

⁢

@

-

[

u

⁢

l

]

⁢

\ar

⁢

@

-

[

u

⁢

l

⁢

l

]

⁢

&

⁢

V

x

If one of the trees is not a

T

⁢

K

1

,

4

it has exactly two vertices of degree 3, which means that

K

is a topological minor (

http://planetmath.org/subdivision

) of

K

3

,

3

.

K

now looks like this. The big circles and

triangles

are presenting different sets of vertices.

\xymatrix

⁢

&

△

&

○

\ar

⁢

@

.

[

l

]

△

\ar

⁢

@

.

[

u

⁢

r

⁢

r

]

⁢

&

⁢

&

⁢

&

○

\ar

⁢

@

.

[

l

⁢

l

⁢

l

]

⁢

\ar

⁢

@

.

[

u

⁢

l

⁢

l

]

∙

\ar

⁢

@

-

[

u

]

⁢

&

∙

\ar

⁢

@

-

[

u

⁢

u

]

⁢

&

∙

\ar

⁢

@

-

[

u

⁢

u

]

⁢

&

∙

\ar

⁢

@

-

[

u

]

⁢

&

⁢

T

x

≠

M

⁢

K

1

,

4

⁢

&

○

\ar

⁢

@

-

[

u

⁢

l

]

⁢

\ar

⁢

@

-

[

u

]

⁢

&

△

\ar

⁢

@

-

[

u

]

⁢

\ar

⁢

@

-

[

u

⁢

r

]

⁢

\ar

⁢

@

-

[

l

]

⁢

&

⁢

V

x

Thus if

G

⪰

K

5

then

G

contains a topological minor (

http://planetmath.org/subdivision

) of

K

3

,

3

or

K

5

. Which proofs this theorem.

Title

equivalence

between the minor and topological minor of

K

5

or

K

3

,

3

Canonical name

EquivalenceBetweenTheMinorAndTopologicalMinorOfK5OrK33

Date of creation

2013-03-22 17:47:15

Last modified on

2013-03-22 17:47:15

Owner

jwaixs (18148)

Last modified by

jwaixs (18148)

Numerical id

11

Author

jwaixs (18148)

Entry type

Proof

Classification

msc 05C83

Classification

msc 05C10

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