I Showing that inverse of an isomorphism is an isomorphism

1,456
44
Let ##G## and ##H## be groups, and let ##\phi : G \to H## be an isomorphism. I want to show that ##\phi^{-1} : H \to G## is also an isomorphism. First, note that ##\phi^{-1}## is clearly a bijection as ##\phi## is its inverse. Second, let ##a,b \in H##. Since ##\phi## is surjective, there exist ##x,y \in G## s.t ##a = \phi(x)## and ##b = \phi(y)##. Then ##\phi^{-1}(ab) = \phi^{-1}(\phi(x) \phi(y)) = \phi^{-1}(\phi(xy)) = xy = \phi^{-1} (\phi (x)) \phi^{-1} (\phi(y)) = \phi^{-1}(a) \phi^{-1}(b)##.

Here is my question, why did I only have to use the fact that ##\phi## is surjective and a homomorphism in showing that ##\phi^{-1}## is a homomorphism? Why didn't I have to use that it is injective?
 

fresh_42

Mentor
Insights Author
2018 Award
11,438
7,868
Let ##G## and ##H## be groups, and let ##\phi : G \to H## be an isomorphism. I want to show that ##\phi^{-1} : H \to G## is also an isomorphism. First, note that ##\phi^{-1}## is clearly a bijection as ##\phi## is its inverse. Second, let ##a,b \in H##. Since ##\phi## is surjective, there exist ##x,y \in G## s.t ##a = \phi(x)## and ##b = \phi(y)##. Then ##\phi^{-1}(ab) = \phi^{-1}(\phi(x) \phi(y)) = \phi^{-1}(\phi(xy)) \stackrel{(*)}{=} xy \stackrel{(*)}{=} \phi^{-1} (\phi (x)) \phi^{-1} (\phi(y)) = \phi^{-1}(a) \phi^{-1}(b)##.

Here is my question, why did I only have to use the fact that ##\phi## is surjective and a homomorphism in showing that ##\phi^{-1}## is a homomorphism? Why didn't I have to use that it is injective?
You did use it. Let's assume ##\phi ## is only surjective and we have ##\phi(u)=\phi(v)##. Then how could we justify ##u=\phi^{-1}(\phi(u))=\phi^{-1}(\phi(v))=v## which you used at the equations I marked with ##(*)\,?## So we actually used ##\phi(u)=\phi(v) \Longrightarrow u=v## which is precisely injectivity.
 

fresh_42

Mentor
Insights Author
2018 Award
11,438
7,868
One can define injectivity and surjectivity only by means of morphisms. Let's take the example above: ##\phi\, : \,G \longrightarrow H##.

Then ##\phi ## is injective, if and only if for any functions ##\varphi , \psi \, : \, K \longrightarrow G## from a set ##K## with ##\phi \varphi = \phi \psi## follows ##\varphi = \psi \,.##

And ##\phi ## is surjective, if and only if for any functions ##\varphi , \psi \, : \, H \longrightarrow L## to a set ##L## with ##\varphi \phi = \psi \phi## follows ##\varphi = \psi \,.##

If you like you can show the equivalence of these definitions to the usual ones as an exercise. So injectivity is left cancellation and surjectivity right cancellation. For an isomorphisms we need, resp. have both. That's why I said in an earlier thread, that both directions are needed: ##\phi \phi^{-1} = \operatorname{id}_H## and ##\phi^{-1} \phi = \operatorname{id}_G##.
 

Math_QED

Science Advisor
Homework Helper
1,183
396
You can't talk about inverse functions if the function is not injective...
 

Want to reply to this thread?

"Showing that inverse of an isomorphism is an isomorphism" You must log in or register to reply here.

Related Threads for: Showing that inverse of an isomorphism is an isomorphism

  • Posted
Replies
1
Views
2K
Replies
1
Views
1K
  • Posted
Replies
2
Views
2K
Replies
1
Views
550
Replies
1
Views
2K
Replies
32
Views
20K
  • Posted
Replies
2
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top