Is this map from $(S^1)^n$ (n-copies of $S^1$) into $S^1$ continuous? Let $S^1$ be the $1$-dimensional sphere given by $S^1= \{e^{i\theta} \ | \ 0 \leq \theta < 2 \pi \}$.
Define a map $f:(S^1)^n \longrightarrow S^1$ given by $$f(e^{i \theta_1}, \ldots ,e^{i \theta_n})=e^{i (\theta_1+\ldots +\theta_n)/n}$$
I want to check that if the map $f$ is continuous. Below is gist of a proof that disapproves of this statement using some tools from algebraic topology and properties of $f$. But I want to know is there a intuitive way to check this statement, even if for the case of $n=2$.
Here is the gist of the proof: 
If the map $f$ is continuous, it will induce a homomorphism
$$f_*: \pi_1(S^1)^n \rightarrow \pi_1(S^1),$$ 
since $\pi_1(S^1)^n \cong \pi_1((S^1)^n)$.
Moreover, $f$ satisfies properties like $f(x, \ldots, x)=x$ and $f(x_1, \ldots, x_n)=f(x_{\sigma(x_1)},\ldots,x_{\sigma(x_n)})$ where $\sigma$ is any permutation of $n$ elements, it is easy to check that $f_*$ also satisfies these condition. The map $f_*$ will induce a homomorphism $g:\pi_1(S^1) \rightarrow \pi_1(S^1),$ given by $$g(x)=f_*(x,0,\ldots,0)$$ and $ng(x)=nf_*(x,0,\ldots,0) = f_*(x,\ldots,x)=x$. Thus multplication by $n$ is an automorphism of $\pi_1(S^1)$, which is clearly false since $\pi_1(S^1) \cong \mathbb{Z}$.
 A: You proof is correct, but I think it invokes unnecessary ingredients. A straightforward proof is this:
Let $\zeta^{(k)} = e^{i(2\pi - 1/k)}$. Then $\zeta^{(k)} \to e^{i2\pi} = 1$, but $f(\zeta^{(k)},1,\ldots,1) = e^{i(2\pi - 1/k)/n} \to e^{i2\pi/n} \ne 1 = f(1,\ldots,1)$.
A more general point of view is this: Your definition makes $f(z_1,\ldots,z_n)$ a specially chosen $n$-th root of $z_1\cdot \ldots \cdot z_n$, where "$\cdot$" is complex multiplication. Thus we may ask whether is possible that a continuous  map $f : (S^1)^n \to S^1$ has the property $f(z_1,\ldots,z_n)^n = z_1\cdot \ldots \cdot z_n$. In this form the question does not involve a special choice of $n$-th roots.
The answer is "no". Consider $g : S^1 \to S^1, g(z) = f(z,1,\dots,1)$. This would be a continuous map such that $g(z)^n = z$, i.e. $g$ would be a continuous $n$-th root on $S^1$. It is well-known from complex analysis that this is impossible. However, you can also use your argument based on fundamental groups. In fact, let $h : S^1 \to S^1, h(z) = z^n$. Then $h \circ g = id$, thus $h_* \circ g_* = id$ on fundamental groups. But $h_*$ is multiplication by $n$, thus $n g_*(a) = a$ for all $a \in \pi_1(S^1) \approx \mathbb Z$ which is impssible.
