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I know what it means to be a isomorphism in a given category. But now i want to prove the following statements:

  • Isomorphisms are exactly the bijections in the catgeory of Sets (Sets)
  • Isomorphisms are exactly the bijective homomorphisms in the catgeory of Monoids (Monoids)

Are these facts not trivial because it is already the same definition or do i not see the difficulty in the proofs? Can someone givbe a comment/hint on this?

Thank you :)

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    $\begingroup$ If you define a bijection as "injection and surjection" then there obviously is some work to be done. If you define it as "has a two-sided inverse", not so much. For monoids, you should also prove that the candidate-inverse is actually a morphism of the category, i.e. a monoid homomorphism. $\endgroup$ – Lord_Farin Sep 11 '13 at 9:17
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To use the example of sets: given sets $A$ and $B$, and a function $f:A\to B$ ,

  • $f$ is a bijection if, for every $b\in B$, there is exactly one $a\in A$ such that $f(a)=b$.

  • $f$ is an isomorphism if there is a function $g:B\to A$ such that $g\circ f=\mathrm{id}_A$ and $f\circ g=\mathrm{id}_B$.

Those are obviously not the same definition (though, as the exercise is asking you to prove, they are ultimately equivalent). Your job is to argue that for any bijection $f$, there is a function $g$ with the desired properties, and conversely, that any isomorphism $f$ must be a bijection.

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  • $\begingroup$ Oh yes right :) $\endgroup$ – Contravariant Sep 11 '13 at 9:21
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    $\begingroup$ Yes, you have to construct a function $g$ that is both a right- and a left-inverse for $f$ (it is not sufficient to prove that $f$ has a right-inverse, and also that it has a left-inverse, without also proving that they are the same function). $\endgroup$ – Zev Chonoles Sep 11 '13 at 9:32
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    $\begingroup$ I did it on this way: We define $g:Y\rightarrow X$ elementwise: Suppose $y_0\in Y$ then by sujectivity of $f$ we have $x_0\in X$ such that $f(x_0)=y_0$ this $x_0$ is unique because $f$ is injective. Define then $g(y_0)=x_0$. Then we can prove $f\circ g=id_Y$ and $g\circ f=id_X$. Right?! $\endgroup$ – Contravariant Sep 11 '13 at 9:35
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    $\begingroup$ Yes, that's right. That shows that any bijection is an isomorphism. Now you have to show the other direction: that any isomorphism is a bijection. $\endgroup$ – Zev Chonoles Sep 11 '13 at 9:38
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    $\begingroup$ I think you should keep trying. I would also appreciate it if you did not use an exclamation point with your questions - there is no need to yell. $\endgroup$ – Zev Chonoles Sep 11 '13 at 9:43

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