Isomorphism and Binary operation

[Gale]

Homework Statement

(i) If $(X,*)$ is a binary operation, show that the identity function
$Id_X : X \rightarrow X$is an isomorphism.

(ii) Let $(X_1, *_1) and (X_2, *_2)$ be two binary structures and let $f : X_1 \rightarrow X_2$ be an isomorphism of the binary structures. Show that $f^-1 : X_2 \rightarrow X_1$ is also an isomorphism.

(iii) Let $(X_1, *_1), (X_2, *_2), (X_3, *_3)$ be three binary structures and
let $f : X_1 \rightarrow X_2$ and $g : X_2 \rightarrow X_3$ be isomorphisms of the binary structures. Show that $g \circ f : X_1 \rightarrow X_3$ is also an isomorphism.

(iv) Denote the statement that $(X_1,*_1)$ and $(X_2, *_2)$are isomorphic by $(X_1, *_1) \cong (X_2, *_2)$. Using the above, show that $\cong$ is reflexive, symmetric and transitive.

Homework Equations

The Attempt at a Solution

Okay, so I'm a bit confused with how to work with isomorphisms and binary operations in general. I'm don't know how to approach the first half of the problem, so I can't really do the rest either. Am I supposed to choose elements from the set X and work my proofs from there, or is there some other approach I should be taking? Besides that, I'm not sure I entirely understand the more general premise of the problems:

Starting with i) I'm not sure why the identity function is only isomorphic when there exists a binary relation. I'm not very confident in my understanding, but it seems like the identity function would always be isomorphic?

ii) I'm not sure how to start a proof for this, but since f is an isomorphism, isn't it necessarily bijective so obviously it would have an inverse? I'm confused as to what the proof is supposed to prove?

 

[damabo]

the definition of an isomorphism is that the function is bijective and linear. I think the first question pertains to the identical function, which does not (?) make use of * .

 

[Gale]

the definition of an isomorphism is that the function is bijective and linear. I think the first question pertains to the identical function, which does not (?) make use of * .

This is why I don't understand the problem... I think. The identity function isn't ismorphic with itself, because its only operating on the set X, and isomorphisms are from functions to functions? But I suppose the question could be relating the Id function to the initial binary operation? But if that's the case, I'm still not sure where to start the proof for that...

 

[Gale]

Okay, so I've been trying to figure out part i, does this make sense:

If $Id_x$ is isomorphic then $Id_x (x_1 * x_2)= Id_x(x_1)* Id_x(x_2)$
$$Id_x(x_1 * x_2)= x_1 * x_2; Id_x (x_1)= x_1, Id_x(x_2)= x_2 \Rightarrow Id_x(x_1) * Id_x(x_2)= x_1 *x_2= Id_x(x_1*x_2)$$

Therefore $Id_x$ is isomorphic?

For part ii) I was trying to use a similar argument, but I'm not sure it works. Is there a different way to approach it?

 

[micromass]

Okay, so I've been trying to figure out part i, does this make sense:

If $Id_x$ is isomorphic then $Id_x (x_1 * x_2)= Id_x(x_1)* Id_x(x_2)$
$$Id_x(x_1 * x_2)= x_1 * x_2; Id_x (x_1)= x_1, Id_x(x_2)= x_2 \Rightarrow Id_x(x_1) * Id_x(x_2)= x_1 *x_2= Id_x(x_1*x_2)$$

Therefore $Id_x$ is isomorphic?

That is correct.

For part ii) I was trying to use a similar argument, but I'm not sure it works. Is there a different way to approach it?

Part (ii) is a bit tricky. You'll need to prove that for $y_1,y_2\in X_2$, that $f^{-1}(y_1*y_2)=f^{-1}(y_1)*f^{-1}(y_2)$. What does $f^{-1}(y_1),~f^{-1}(y_2),~f^{-1}(y_1*y_2)$ mean by definition?

 

[UNChaneul]

That is correct.



Part (ii) is a bit tricky. You'll need to prove that for $y_1,y_2\in X_2$, that $f^{-1}(y_1*y_2)=f^{-1}(y_1)*f^{-1}(y_2)$. What does $f^{-1}(y_1),~f^{-1}(y_2),~f^{-1}(y_1*y_2)$ mean by definition?

Technically he just showed $Id_{x}$ is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that $Id_{x}$ is also bijective, granted that follows rather trivially from the definition of the identity function, but I think he should still state something like "And $Id_{s}$ is bijective by definition" after that bit.

 

[micromass]

Technically he just showed $Id_{x}$ is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that $Id_{x}$ is also bijective, granted that follows rather trivially from the definition of the identity function, but I think he should still state something like "And $Id_{s}$ is bijective by definition" after that bit.

Yes, you are correct! It should have been added.

 

[Gale]

Technically he just showed $Id_{x}$ is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that $Id_{x}$ is also bijective, granted that follows rather trivially from the definition of the identity function, but I think he should still state something like "And $Id_{s}$ is bijective by definition" after that bit.

Thanks... someone else noticed that today as well. I suppose I wasn't solid on the definition of an isomorphism... so I didn't realize I needed a bijection. We haven't spent any time discussing other morphisms, so I didn't understand the difference.

(Also, I'm a she, not he)

 

[UNChaneul]

Thanks... someone else noticed that today as well. I suppose I wasn't solid on the definition of an isomorphism... so I didn't realize I needed a bijection. We haven't spent any time discussing other morphisms, so I didn't understand the difference.

(Also, I'm a she, not he)

English needs a pronoun for when one does not know gender xD. Using "they" sounds so awkward when referring to one person!