I read the definition of a universal property of a product of topological spaces. Generally it says that for map $f$ from $\mathbb{X}$ to $\mathbb{Y}$ and map $g$ from $\mathbb{X}$ to $\mathbb{Z}$ there exist unique map $h$ from $\mathbb{X}$ to $\mathbb{Y}\times \mathbb{Z}$ such that projection of image of $h$ equals to image of $f$ and image of $g$. For what purpose we use this property and isn't it just $h(x)=(f(x),g(x))$ ?
Thanks.
It sounds like the way it was presented makes it seem tautological. Let's put it slightly differently.
Any topological space $Z$ equipped with maps $p:Z\to X$ and $q:Z\to Y$ has the universal property of being the product of the spaces $X$ and $Y$ if and only if for every pair of maps $f: W\to X$ and $g:W\to Y$ there exists a unique map $h: W\to Z$ such that $p\circ h=f$ and $q\circ h = g$.
You are completely correct that choosing $Z=X\times Y$ with $p=\pi_1$ and $q=\pi_2$ gives you something that satisfies the universal property and indeed will lead to $h(x)=(f(x),g(x))$. But we could also choose $Z=Y\times X$ with $p=\pi_2$ and $q=\pi_1$ for which $h(x)=(g(x),f(x))$. So the universal property does not (in general) uniquely identify a topological space. However, as you should prove, given two topological spaces each equipped with a pair of "projections", if both satisfy the universal property then they are homeomorphic, in fact, there exists exactly one homeomorphism that takes the "projections" of one to the "projections" of the other.
This last fact (that actually holds, suitably formulated, for all "universal properties") is very useful. Given any topological space $T$ and pair of functions $T\to X$ and $T\to Y$, if we can show that $T$ with those functions satisfies the universal property of products, then we know that $T\cong X\times Y$. $T$ may not always be obviously a set of pairs with the product topology.