Proving f(f⁻¹(B)) = B for All B in Y

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Homework Statement


Prove that if f: X \rightarrow Y is onto, then f(f^{-1}(B))=B \forall B \in Y

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The Attempt at a Solution

 
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What does it mean for a function to be onto? What kind of inverse does f possesses iff it is onto?
 
Onto means that for a function f:A \rightarrow B if \forall b \in B there is an a \in A: f(a)=b

The inverse means that if you take the f^{-1}(b) that it should map back to a?
 
Correct. But note that a right inverse exists if the function is onto. I.e., if g is a right inverse of f, then f(g(y)) = y, for every y in Y. What you need to prove is a direct consequence of this fact. (I used "g" rather than "f^-1" for the right inverse to avoid confusion leading to a conclusion that f^-1 is an inverse, i.e. both left and right).
 
So I need to prove that if f(y)=Y and f^{1}(Y)=y, that f(f^{1}(Y))=Y?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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