Question: I'm trying to prove this rather simple fact that the following mapping is a covering and to check it's rank

$p\colon S^1\times S^1\rightarrow S^1\times S^1$; $p(z\_1,z\_2)=(z\_1^2,z\_2^3).$

So I think that rank is 6 - $p^{-1}(a,b)$, for every two $z\_1$ that would give $a$ after taking a square, there are three $z^2$ that would give $b$ - so there are 6 such points.

1. $p$ being a surjection is obvious. Being a covering also, but I'm getting stuck trying to give a precise argument.. How to formulate it?
2. Does every $f\colon\mathbb{RP}^2\rightarrow S^1\times S^1$ lift with respect to this $p$? I guess that yes, and maybe it doesn't even depend on the domain - projective space. It suffices to take 
$(e^{i\alpha /2},e^{i\beta /3})$, where $(e^{i\alpha},e^{i\beta})$ is the the image of $f$,
as the image of $h\colon\mathbb{RP}^2\rightarrow S^1\times S^1$, and then $ph=f$. I hope it's clear what I want to say. Is that correct?

Thanks

Answer: 
Your idea is wrong on two counts:

1. You do not cover the case when you approach $c$ from a "mixed" sequence, i.e. from a sequence that contains both rational and irrational elements. There is nothing to show that you can simply ignore those cases.
2. You are trying to prove that $f$ is differentiable at $c$, but this may not be the case. For example, if $g(x)=h(x)=|x|$, then $f$ is **not** differentiable at $0$!

---

I suggest you prove the continuity of $f$ at $c$ using the $\epsilon$-$\delta$ definition of continuity. You should, for the most part, only require the definition of what it means that $g$ (and $h$) is continuous at $c$, and everything should fall right out:

You need to prove that:

> 
> For every $\epsilon > 0$ there exists a $\delta>0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$.
> 
> 
> 

You already know that:

1. For every $\epsilon > 0$ there exists a $\delta\_1>0$ such that $|x-c|<\delta\_1$ implies $|g(x)-g(c)|<\epsilon$.
2. For every $\epsilon > 0$ there exists a $\delta\_2>0$ such that $|x-c|<\delta\_2$ implies $|h(x)-h(c)|<\epsilon$.

You also know that $|f(x)-f(c)|$ is equal to either $|g(x)-f(c)|$ **or** $|h(x)-h(c)|$.
To show that the limit of $f$ in $c$ is $f(c)$ you need to show that for every $\epsilon > 0$ there's $\delta > 0$ such that for every $x\in (c-\delta , c+\delta)$ you have that $|f(x)-f(c)|<\epsilon$.

Since $g$ and $h$ are both continuous at $c$ and equal $f(c)$ in that point, you have that for $\epsilon > 0$ you have $\delta \_1 >0$ such that for $x\in (c-\delta \_1 , c+\delta \_1)$ it holds that $|g(x)-f(c)| <\epsilon$ and $\delta \_2 >0$ such that for $x\in (c-\delta \_2 , c+\delta \_2)$ it holds that $|h(x)-f(c)| <\epsilon$.

Now by taking $\delta = \min\{\delta \_1,\delta \_2\}$ you get what you want.