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In completion to my question here:

Proving that $f^{-1} \in \operatorname{Hom}(Y,X).$

I know that: If $X,Y$ are groups and $f\in \operatorname{Hom}(X,Y)$ is bijective, then $f^{-1} \in \operatorname{Hom}(Y,X).$

My question is:

Why this statement is not correct in other categories than groups? could anyone help me answer this question, please?

In helping me answering my previous question @Tsemo in the previous question asked me that question:

how we define the fact that $Hom_{C}(X,Y)$ is bijective for category $C,$could you please help me in answering that question? I have no clear definition in my mind.

EDIT:

My confusion arises from the definition of Isomorphism that my professor gave to us, he said:

$f\in \operatorname{Hom}(X,Y)$ is as isomorphism if it is bijective and $f^{-1} \in \operatorname{Hom}(Y,X).$

He added that: in the category of groups $f\in \operatorname{Hom}(X,Y)$ is as isomorphism if it is bijective only.

This what confuses me, because I used to know that Isomorphism means homomorphism and bijection.

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    “Bijective” doesn’t really make sense in a category. “Isomorphism” does, in which case the result follows directly from the axioms. – Randall Sep 23 '20 at 00:50
  • I do not understand you @Randall could you please explain more details? –  Sep 23 '20 at 00:54
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    What constitutes a morphism is category dependent. For example, suppose morphisms were differentiable functions. Then $f(x)=x^3$ is a differentiable bijection whose inverse is not a morphism. – John Douma Sep 23 '20 at 00:55
  • Do you have an answer to my main question @Randall? sorry for my stupidness. –  Sep 23 '20 at 00:55
  • In the category of topological spaces, there are many bijective continuous functions whose inverse is not continuous. Have a look at https://math.stackexchange.com/questions/378717/finding-counterexamples-bijective-continuous-functions-that-are-not-homeomorphi – Julian Rosen Sep 23 '20 at 00:58
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    There is no definition of bijective in an abstract category. Not all categories have functions as morphisms. – Randall Sep 23 '20 at 00:58
  • @JulianRosen Yeah I know that, but we are here speaking about homomorphism not continuity ..... I do not see any relation between a function being homomorphism and it is being continous. –  Sep 23 '20 at 01:04
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    The morphisms in the category of topological spaces are continuous functions. – Julian Rosen Sep 23 '20 at 01:12
  • @JohnDouma that make sense ..... can you give me an example whose inverse is not homomorphism? –  Sep 23 '20 at 01:12
  • that make sense ..... can you give me an example whose inverse is not homomorphism?@JulianRosen –  Sep 23 '20 at 01:13
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    The question that I linked gives a couple examples of continuous bijections whose inverse is not continuous. One example: take $Y$ to be any non-discrete space, let $X$ have the same underlying set as $Y$ but with the discrete topology, and let $f:X\to Y$ be the identity function. – Julian Rosen Sep 23 '20 at 01:18
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    @Smart20 In category theory we write $\textrm{Hom}(X, Y)$ for the set of morphisms from $X$ to $Y$ without regard to whether or not such morphisms are called homomorphisms or not. – Zhen Lin Sep 23 '20 at 03:31
  • @ZhenLin ohhh that removes my confusion ..... thanks alot Zhen –  Sep 23 '20 at 03:33

1 Answers1

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In a category $\mathcal C$, a morphism $f:X\to Y$ is called an isomorphism if there is a morphism $g:Y\to X$ such that $g\circ f = 1_X$ and $f\circ g = 1_Y$.

If your category is concrete, i.e.

  • objects are sets with extra stuff (e.g. groups, spaces),
  • morphisms are stuff-preserving functions (e.g. homomorphisms, continuous functions),

then, unwrapping the definition, a morphism $f:X\to Y$ is an isomorphism iff it has an inverse ($f^{-1}:Y\to X$) preserving the extra stuff ($f\in\hom(Y,X)$, i.e. not just a bijection).

Your teacher proved the following:

Fact. In the category of groups and homomorphisms between them, a (homo)morphism $f:X\to Y$ is an isomorphism iff its a bijection. In other words, the set-theoretic inverse $f^{-1}:Y\to X$ is already a homomorphism.

This does not occur in general:

Fact. In the category of spaces and continuous functions, there are bijections which are not homeomorphisms. For instance, if $X$ has nonequivalent topologies $\mathcal O_1, \mathcal O_2$ then the identity $id:(X,\mathcal O_1)\to (X,\mathcal O_2)$ is a bijection, but not an homeomorphism. Example.

Fact. In the category of smooth manifodls and smooth functions, there are bijections which are not diffeomorphisms. John Douma's $f(x) = x^3$ is a counterexample.

Daniel Teixeira
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