How can I prove a function is surjective? In the function $f: \Bbb{R}\to \Bbb{R}$, $$f(x) = 4x+7$$ we take $x = y-\frac{7}{4}$ and show that $f(x)=y$. How can this method prove that this function is surjective. Could someone explain this please?
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Because for any $y$ you can find an $x$ such that $f(x)=y$. – 3x89g2 May 20 '15 at 06:15
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To prove that something is surjective we need to show that every element in your co-domain gets hit so one way to show this is to show that you have a right inverse the reason or the intuition to as why this is true is as follows when you do $f(f^{-1}(y)) = y$ what your doing essentially is going back first by inverse into your domain from that specific element x that should map to y and then go back to that original y again and so if we didn't have our map as being surjective what would happen is the right inverse wouldn't be defined since we wouldn't have anything that gets mapped to y So that is why it works. Actually in your case $f^{-1}(x) = \frac{x - 7}{4}$ will be your inverse your function has a left and a right inverse so is bijective function ! But what your doing when your proving that something is surjective is your proving that it has a right inverse essentially.
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In order to show $f$ is surjective, we need to show that for all $y\in \Bbb{R}$, there is some $x\in \Bbb{R}$ such that $f(x)=y$. But this is precisely what is done here, although it should be $x=1/4y-7/4$, since for any $y$, $f(\frac{1}{4}y-\frac{7}{4})=y-7+7=y$.
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