n fold symmetric product topology and universal property Let $X$ be a topological space, and $n$ a natural number. Define Sym $^{n} X,$ the $n$ -fold symmetric product of $X,$ to be the set of equivalence classes of $n$ -tuples $\left(x_{1}, \ldots, x_{n}\right)$ under the relation
$$
\left(x_{1}, x_{2}, \ldots, x_{n}\right) \sim\left(x_{\pi(1)}, x_{\pi(2)}, \ldots, x_{\pi(n)}\right)
$$
whenever $\pi$ is a permutation of $\{1, \ldots, n\} .$ There is a surjective function $q: X^{n} \rightarrow \operatorname{Sym}^{n} X$ taking an $n$ -tuple to its equivalence class. Endow Sym $^{n} X$ with the quotient (co-induced) topology.
Give $\mathbf{C}$ the usual topology. Show that the continuous $\operatorname{map} f: \mathbf{C}^{2} \rightarrow \mathbf{C}^{2}$ given by $f(x, y)=(x+y, x y)$ factors as

for some continuous bijection $h$
My strategy:
Obviously, if $(a,b)\sim (c,d)$, then $f(a,b)=f(c,d)$. Thus, by the universal property of $\text{Sym}^2C$, there exists a unique continuous function $h:\text{Sym}^2C\rightarrow C^2$ s.t. $f=h\circ q$.
But how do we show $h$ is bijective?
 A: This is not really a matter of topology. This is a purely algebraic fact about algebraically closed fields. In fact, the injectivity holds for arbitrary domains.
To show injectivity, you need to show that a pair of complex numbers is determined by it sum and product.
If you write $s$ and $p$ for the sum and product, respectively, then the elements of the pair satisfy $x+y=s$ and $xy=p$, so each element satisfies $x(s-x)=p$. Now, this equation has at most two possible solutions $x$, and each yields the corresponding $y$. If there are two solutions, then choosing the other one amounts to swapping $x$ and $y$, so the unordered pair remains the same.
For surjectivity, you need to show that given $s$ and $p$, you can find an $x$ such that $p=x(s-x)$, i.e. that the equation $x^2-sx+p=0$ has a solution. But this is clearly true for $\mathbf C$ (as it is for any algebraically closed field, or simply any field with no quadratic field extensions).
(In case you are curious, to obtain a general statement for larger symmetric products, you should use elementary symmetric polynomials in place of $x+y$ and $xy$.)
A: Define $h$ first by $h([(x,y)] = f(x,y)$; there is no choice, because of the factorisation property.
Suppose that $h([(x,y)])=h([(x',y')])$. So $x+y=x'+y'$ and $xy = x'y'$. Now show that $x=x'$ and $y=y'$ or $x=y'$ and $y=x'$, which in both cases means $(x,y) \sim (x',y')$ so $[(x,y)]= [(x',y')]$.

 Note that $x,y$ are the unique two roots of $p(t):=(t-x)(t-y)=t^2+at+b=0$ (as a polynomial in $t$, so the coefficients are $a=x+y$ and $b=xy$) and the corresponding polynomial for $x',y"$ is the same $p(t)$ (!). By uniqueness up to order of the roots of a square polynomial $\{x,y\}=\{x',y\}$; both are equal to the set $\{t\mid p(t)=0 \}$  etc.

Surjectivity is quite simple, just note that $f$ is, and so $h$ is too.
