Is this composition bijective?

**Sheld** · October 12th 2011, 10:53 AM

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

then $f \;o \; g \;o f^{-1}$ 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 $g:X-X$ to a unique function, that goes from $Y->Y)$

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.

**Plato** · October 12th 2011, 12:34 PM

Originally Posted by Sheld

If $f:X->Y$ is bijective and $g$ is any map s.t $g:X->X$
then $f \;o \; g \;o f^{-1}$ is bijective.

Let $X=Y=\mathbb{R}$ and $f(x)=x~\&~g(x)=x^2$ .

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

**Deveno** · October 12th 2011, 01:13 PM

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.

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