Claim: Let $f: \mathbb{R^+} \to \mathbb{R}$ defined by $f(x)=\sqrt{x}$. Then $f$ is an injection Show that the function is an injection by finding a left-inverse of the function, or explain why you can't find such a function given $f$.
Let $f: \mathbb{R^+} \to \mathbb{R}$ defined by $f(x)=\sqrt{x}$.
My Attempt
Proof. To prove that $f$ is injective, we need only to show that there exists a function $g: \mathbb{R} \to \mathbb{R^+}$ such that for all $x\in \mathbb{R^+}$ we have $(g \circ f)(x)=x$.
Let $g: \mathbb{R} \to \mathbb{R^+}$ defined by $g(x)=x^2$.

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*Totality: Because every element of the domain, $\mathbb{R}$, is assigned to something; that is, the square of any real number is defined, it follows that the totality condition is satisfied.


*Existence: Every assignment is not a well-defined existing element of $\mathbb{R^+}$. Counterexample: Let $x=0 \in\mathbb{R}$, but $g(0) = 0 \notin \mathbb{R^+}$


*Uniqueness: As every element of the domain is assigned to only one element of the codomain $\Rightarrow$ existence is satisfied.
As $g$ does not satisfy the condition $\forall x \in \mathbb{R}$, $\exists! y \in \mathbb{R^+}$ such that $g(x)=y$ $\Rightarrow$ $g$ is not a well-defined left-inverse of $f$.
However, if $f: \mathbb{R^+} \to \mathbb{R^+}$ defined by $f(x)=\sqrt{x}$, then $f$ is an injection whereby $g: \mathbb{R^+} \to \mathbb{R^+}$ defined by $g(x)=x^2$ is its left-inverse; that is, $\forall x \in \mathbb{R^+}$, $g(f(x)) = x$.
 A: You have only shown that $\textbf{one particular}$ function $g$ does not match the property. To show that $f$ is not injective, you must show that there $\textbf{cannot exists any}$ such $g: \mathbb R \to \mathbb R^+ $. There might be (and there are) other functions $g$ that act as a left-inverse.
Just because YOU don't speak French, it does not mean no one else can speak it either.
Consider $g:\mathbb R \to \mathbb R^+$ by
$$g(x) = \begin{cases}x^2 &\text{ if } x\neq 0 \\ 42 &\text{ if } x=0\end{cases}$$
Then for all $x\in \mathbb R^+$ you have $f(x) \in \mathbb R^+$ so $f(x) \neq 0$ therefore $$(g\circ f)(x) = g(f(x)) = (\sqrt{x})^2 = x$$
and therefore $f$ is injective.
A: Since the OP's analysis has received a response, it is open season to use an alternative method to demonstrate that $f:\Bbb{R}^+ \to \Bbb{R}^+ ~: ~f(x) = \sqrt{x}$ is injective.
$f'(x) = \frac{1}{2\sqrt{x}} > 0 ~$ for all $x > 0.$
Therefore, $f(x)$ is a strictly increasing function.  Therefore, if $x < y$, then $f(x) < f(y)$.  Therefore, $f(x)$ is an injection.
