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Exercise 32 - Chapter 3

edited June 2018 in Exercises

Let \(G\) be a graph, and let \(\textbf{Free}(G)\) be the corresponding free category. Somebody tells you that the only isomorphisms in \(\textbf{Free}(G)\) are the identity morphisms.

Is that person correct?

Why or why not?

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Comments

  • 1.
    edited May 2018

    I presume that the person is making a statement about graphs generally and not a specific graph.

    The person is confused about the difference between a round-trip vs. a trip for which has a return flight.

    Comment Source:I presume that the person is making a statement about graphs generally and not a specific graph. The person is confused about the difference between a round-trip vs. a trip for which has a return flight.
  • 2.
    edited May 2018

    Yes, the only isomorphisms in the free category of any graph are the identity morphisms. This follows from the fact that the morphisms in \(\bf{Free}(G)\) do not obey any equations (other than those required in the definition of a category, i.e., the left and right unit laws).

    Suppose, by way of contradiction, that \(f:A\to B\) is a non-identity isomorphism in \(\bf{Free}(G)(A,B)\). Then there is a morphism \(g:B\to A\) such that \(f\circ g=\text{id}_A\) and \(g\circ f=\text{id}_B\). However, since \(f\) is not an identity morphism and the only equations among elements of \(\bf{Free}(G)(A,B)\) are the left and right unit laws, no such morphism \(g\) exists. In particular, if \(A\neq B\), then there are no equations at all between elements of \(\bf{Free}(G)(A,B)\). Even if \(A=B\), the only option is \(g=\text{id}_A\), which yields \(f\circ\text{id}_A=\text{id}_A\), so that \(f=\text{id}_A\), which contradicts the assumption that \(f\) is not the identity morphism.

    Comment Source:Yes, the only isomorphisms in the free category of any graph are the identity morphisms. This follows from the fact that the morphisms in \\(\bf{Free}(G)\\) do not obey any equations (other than those required in the definition of a category, i.e., the left and right unit laws). Suppose, by way of contradiction, that \\(f:A\to B\\) is a non-identity isomorphism in \\(\bf{Free}(G)(A,B)\\). Then there is a morphism \\(g:B\to A\\) such that \\(f\circ g=\text{id}_A\\) and \\(g\circ f=\text{id}_B\\). However, since \\(f\\) is not an identity morphism and the only equations among elements of \\(\bf{Free}(G)(A,B)\\) are the left and right unit laws, no such morphism \\(g\\) exists. In particular, if \\(A\neq B\\), then there are no equations at all between elements of \\(\bf{Free}(G)(A,B)\\). Even if \\(A=B\\), the only option is \\(g=\text{id}_A\\), which yields \\(f\circ\text{id}_A=\text{id}_A\\), so that \\(f=\text{id}_A\\), which contradicts the assumption that \\(f\\) is not the identity morphism.
  • 3.
    edited June 2018

    I see, some care must be taken when constructing a graph suitable for construction of a Free category.

    When parallel paths are present in the graph the are assumed to be not equal unless there is an equation specifically stating otherwise.

    The only exception to this rule is the case of the identity morphisms which have the implicit \( id \circ id = id \) equation.

    Comment Source:I see, some care must be taken when constructing a graph suitable for construction of a Free category. When parallel paths are present in the graph the are assumed to be not equal unless there is an equation specifically stating otherwise. The only exception to this rule is the case of the identity morphisms which have the implicit \\( id \circ id = id \\) equation.
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