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 $0$!

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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** $|h(x)-h(c)|$.
You started off correctly and then somehow jumped on differentiation which does not seem to make sense here. If we approach $c$ via rational numbers then the limit is taken using the function $g$ and since it is continuous at $c$ the limit is $g(c)$. Similarly when we approach $c$ using irrational values then the limit is $h(c)$ and since $g(c) = h(c)$ you can see that the limit of $f$ as $x \to c$ is same whether we approach via rationals or irrationals.

Next we need to ensure that the limit is also equal to $f(c)$. For this we need to check the case when $c$ is rational and the case when $c$ is irrational. Clearly if $c$ is rational then $f(c) = g(c)$ and hence the limit is equal to $f(c)$. Similarly we deal with the case when $c$ is irrational.

A much shorter/better proof goes like this. Let us first take the case when $c$ is rational. Then $f(c) = g(c)$. Now both $g, h$ are continuous at $c$ and $g(c) = h(c) = f(c)$ hence we can get all the values of $g(x)$ and $h(x)$ to be arbitrarily close to $f(c)$ by taking $x$ sufficiently close to $c$. Since the values of $f(x)$ are basically values of $g(x), h(x)$ depending or the rationality/irrationality of $x$ it follows that values of $f(x)$ can be made arbitrarily close to $f(c)$ by choosing $x$ sufficiently close to $c$. This means that $f$ is continuous at $c$. Same argument works when $c$ is irrational.