1. Is this composition bijective?

If $\displaystyle f:X->Y$ is bijective and $\displaystyle g$ is any map s.t $\displaystyle g:X->X$

then $\displaystyle f \;o \; g \;o f^{-1}$ is bijective.

the compostion is a function that goes from Y to Y. I am pretty sure it's onto, since f is onto. It's injective because g uniquely determines what the composition ends up as.

(we are sending $\displaystyle g:X-X$ to a unique function, that goes from $\displaystyle Y->Y)$

I think the real question I'm asking is, does this define an isomorphism?

Sorry if this is unclear. I tried my best to word it coherently. I will try to clarify if needed.

2. Re: Is this composition bijective?

Originally Posted by Sheld
If $\displaystyle f:X->Y$ is bijective and $\displaystyle g$ is any map s.t $\displaystyle g:X->X$
then $\displaystyle f \;o \; g \;o f^{-1}$ is bijective.
Let $\displaystyle X=Y=\mathbb{R}$ and $\displaystyle f(x)=x~\&~g(x)=x^2$.

$\displaystyle f\circ g\circ f^{-1}(-1)=f\circ g\circ f^{-1}(1)$, bijective?

3. Re: Is this composition bijective?

suppose g(x1) = g(x2) for x1 ≠ x2 in X.

since f is bijective, f^-1 is bijective, thus surjective.

so we can find y1,y2 in Y with f^-1(y1) = x1, f^-1(y2) = x2.

note that y1 ≠ y2 (since f^-1 exists, it is a function).

now fogof^-1(y1) = f(g(f^-1(y1))) = f(g(x1)) = f(g(x2)) = f(g(f^-1(y2))) = fogof^-1(y2),

so the composition is not injective, so it can't be bijective, either.