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Let $Z = \{(a, b) : a \ge 0 \text{ and } b = 0, \text{ or } a = 0 \text{ and } b \ge 0\}$.

Find a $C^\infty$ smooth function $\phi : \mathbb{R} \to \mathbb{R^2}$ which is $1$−$1$ and onto $Z$.

Q) Can someone please explain what $\phi$ is?

I think it's a tangent vector to my function evaluated with respect to or dependent on time?

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    $Z$ is basically the $x$ and $y$ axes in the positive directions. An obvious bijection would be $\phi(x) = (x,0)$ for $x \ge 0$ and $\phi(x) = (0,-x)$ for $x < 0$, but is this $\phi$ "smooth"? – 2017-01-11
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    $\phi$ is a function. It's a smooth function, which means that we can take as many derivatives as we'd like. – 2017-01-11
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    Think its $C^0$ smooth what more interested in what phi was then what i was actually being asked >.> – 2017-01-11
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    @Alexis Olson If the [derivative](https://en.wikipedia.org/wiki/Derivative#Derivatives_in_higher_dimensions) of such a function $\phi$ exists then it equals its [Jacobian matrix](https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant). Let $\phi(t):=(\phi_1(t),\phi_2(t))$. The requirement that $\phi$ be $C^{\infty}$ means that we want the derivatives $\frac{d^n\phi_1}{d t^n}$ and $\frac{d^n\phi_2}{d t^n}$ to exist (and be continuous, which is superfluous in this case) for all $n$. In your example, $\phi_1(t)=\max(0,t)$ isn't differentiable at $0$. – 2017-01-11

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The $\phi$ in question needs to be a function. If we think about the input parameter as "time", then $\phi$ defines a trajectory in $\Bbb R^2$. Our function needs to be 1-1, which means we can't pass through the same point twice, and it needs to be onto $Z$, which means the path needs to cover all of $Z$.

A function that almost works, but fails to be differentiable, is $$ f(t) = \begin{cases} (|t|,0) & t < 0\\ (0,t) & t \geq 0 \end{cases} $$ To get a function that will work, we can use any smooth function $g(t)$ such that $g(t) = 0$ when $t<0$, $g(t)$ is increasing when $t \geq 0$, and $g(t) \to \infty$ as $t \to \infty$. With such a function, we may define $$ \phi(t) = (g(-t),g(t)) $$ note that this is exactly the same as defining $$ \phi(t) = \begin{cases} (g(-t),0) & t < 0\\ (0,g(t)) & t \geq 0 \end{cases} $$

One example of such a $g$ is $$ g(t) = \begin{cases} 0 & t \leq 0\\ t e^{-1/t} & t > 0 \end{cases} $$

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    Can you explain a bit about what "differentiable" means for a function that maps $\mathbb R$ to $\mathbb R^2$ and why the first function fails? – 2017-01-11
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    the first function is $C^0$ so its continuous but you cant take the derivative at 0 cause it makes a Hard turn on the graph. – 2017-01-11
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    @Alexis A function from $\Bbb R$ to $\Bbb R^n$ is differentiable iff each of its component functions are differentiable. I won't say more than that. – 2017-01-11