# Proving isomorphisms

1. Feb 10, 2012

### trap101

Prove that if S: U-->V and T: V-->W are isomorphisms, then TS is also an isomorphism. Also What is (TS)^-1

Attempt: So I said since S and T are already isomorphisms, then there exist S^-1 and T^-1 So:

T-1T = (Iv)S = S ==> S-1S = Iu

(TS)S-1 = T(Iv) = TT-1 = Iw

now if this is the correct way of proving it, how does it show that TS is an isomorphism?

Also stuck on the second part.

Cheers

2. Feb 10, 2012

### Dansuer

I guess that by TS you mean the composition of function.

What are the conditions that a function has to satisfies to be an isomorphism?
Then you have to prove TS satisfies those condition and you're done :)

3. Feb 10, 2012

### micromass

Staff Emeritus
This:

and

Could you explain those equalities??

To prove your result, do you have any idea what $(TS)^{-1}$ will be?? (if you know something about matrices, then you could try the same thing with T and S matrices).

4. Feb 10, 2012

### trap101

@Dansuer yes the composition. The conditions to satisfy an isomorphism are that the transformation, I guess in this case TS has to be surjective and injective, as well TS has to be invertible. Problem I'm having is how exactly do I show that? These abstract things always get me and I end up spending huge amounts of time on what should take a few mins...

@micromass: the Iv represent the identity transformation that you would get from multiplying through by the inverse T-1, I think, but maybe i'm not even suppose to do that.

5. Feb 10, 2012

### Dansuer

Ok so an isomorphism is injective and surjective. try to prove this about TS using the fact that T and S are isomorphism and so they are injective and surjective.

There is a third condition that an isomorphism has to satisfies though.

Yeah this abstract stuff confuses me a lot too. when i'm stuck, i write down all the definitions of the object i'm working with. This helps me a lot. Maybe it will work for you too

6. Feb 11, 2012

### Fredrik

Staff Emeritus
The identity transformation on V is the map Iv:V→V defined by Iv(x)=x for all x. So the two equalities that micromass quoted are wrong. Unless you mean something by "T(Iv)" that we don't understand. I'm guessing that you just mean the product (i.e. composition of functions) TIv=T°Iv. To see that the equality is wrong, just try evaluating (TIv)(x)=T(Iv(x))=...

This problem requires you to know that an isomorphism is a linear bijection. So to show that TS is an isomorphism, you must prove that TS is linear and bijective (i.e. injective and surjective), so I suggest you do it in three steps:

1. TS is linear.
2. TS is injective.
3. TS is surjective.

By the way, when people have helped you with a problem, and you're ready to move on to the next problem, it would be nice if you could post a comment about it before you abandon the thread.

7. Feb 11, 2012

### trap101

I'm probably missing something blatantly obvious when it comes to linearity so at the moment I'll say I don't know how to prove it.

As for bijectivity, the only concept that pops to mind that would work, would be to show that the dim ker(TS) = 0 , which would imply injectivity and then imply dim (Im) = n which means it's surjective. But I wrote the question you see above word for word from the text so I wasn't given any specific spaces to work with. So I'm stuck in how to start.

Should I start: Since we know S and T are both isomorphic this means they each individually have the properties of being injective and surjective; this being the case the composition TS would also hold these properties?

@Fredrik apologies mate, i'm still learning the etiquette. Along with the appreciation that math from here on in is not going to be 30 min plug and chug anymore....sigh. But also interesting

8. Feb 11, 2012

### Fredrik

Staff Emeritus

Linearity: TS(ax+by)=...(use the linearity of S and T here)...=aTSx+bTSy

Injectivity: Let x and y be arbitrary members of U such that TS(x)=TS(y). Use the properties of T and S to prove that x=y.

Surjectivity: Let z be an arbitrary member of W. Use the properties of T and S to prove that there's an x in U such that TSx=z.

9. Feb 11, 2012

### trap101

So here are my attempts at the three parts you indicated:

1: Linearity:

TS(aX + bY) = TS(ax1 +by1,.......,axn + byn)
= TS(ax1 + by1,......,axn + byn)
= TS( ax1+ by1),.........TS(axn +byn)
= TS(ax1) + TS(by1),.....,TS(an + byn)
= TS(a(x1,...,xn) + TS(b(y1,...,yn)
= aTS(X) + bTS(Y)

2. Injectivity: If I'm letting TS(x) = TS(y)....since I know the properties of the individual S and T, I know they are isomorphic which implies they have inverses. Would it be as simple as:
T-1TS(x) = S(x), then S-1S(x) = Iv(x) = x.
And do the same steps to the other side but instead get "y"?...doesn't seem like the right thing to do, but that's the hunch I went on.

10. Feb 11, 2012

### Fredrik

Staff Emeritus
There's no need to break the vectors up into components. Think about what the statement "S is linear" means, and use that.

Yes, applying S-1T-1 to both sides gets the job done.

11. Feb 11, 2012

### trap101

Got it. But what does proving that x = y show? Because the definition of injectivity is that whenever f(x1) = f(x2) then x1 = x2......Oh!...I think I see it now......Thanks for having the patience.