I am asked to determine whether a function $f(x) = x^{5} + 1$ is a bijection in $\mathcal{R}$. Proving that the function is one-to-one, I come up with the following:
$\begin{array}{lcl}f(x)& = &f(y)\\x^{5} + 1& = &y^{5} + 1\\x^{5}& = &y^{5}\\x&=&y\end{array}$
However, I am a bit confused about finding whether the function is onto. Based on the definition $\forall x\exists y(f(x) = y)$, I tried the following:
$\begin{array}{lcl}y& = &f(x)\\ y& = &x^{5}+1\\x^5& = &y-1\\x& = &(y-1)^{\frac{1}{5}}\end{array}$
Substituting the right hand side of the equation for $x$ in the above definition for onto functions, I get the following: $\forall x\exists (y-1)^{\frac{1}{5}}f[(y-1)^{\frac{1}{5}}=y]$
However, plugging in various values for $y$, the equation always seems to evaluate to the value for $x$. I came to the conclusion that this was not onto, and therefore not a bijection, but the answer in my book says that it is indeed a bijection. What am I doing incorrectly here?
Edit: Silly mistake -- I was simply substituting the value for $x$ as the function body, not accounting for the original function definition.