I have a question based on following exercise from Bosch's "Algebraic Geometry and Commutative Algebra" (page 461):
(*) 
Let $\mathbb{P}^m$ projective space and $\mathcal{L}:= \mathcal{O}_{P^m}(d)$ an invertible sheaf.
The aim is to construct a map $\varphi:\mathbb{P}^m \to \mathbb{P}^n$ with $n= \binom{m+d}{m}-1$. It is called also the Veronese embedding.
The construction seems quite straight forward: There exist exactly $\binom{m+d}{m}$ monomials $s_{k_0,k_1,..., k_m}:=T_0^{k_0}T_1^{k_1}...T_m^{k_m}$ of degree $d$ (so we have $\sum k_i = d$)
and therefore the data $(\mathcal{L}:=\mathcal{O}_{P^m}(d); s_{k_0,k_1,..., k_m})$ determines such $\varphi$ uniquely via
$\mathcal{O}_{P^m}(d)= \varphi^*\mathcal{O}_{P^n}(1)$ and $s_{k_0,k_1,..., k_m}:=f^{\#}(T_{k_0,k_1,..., k_m})$ for $\mathbb{P}^n= Proj(k[T_{k_0,k_1,..., k_m}]_{k_0+k_1,..., k_m=d})$.
My question is how to show that this map "coinsides" with the "common" Veronese embedding in appropriate sense as defined for projective spaces $\mathbb{P}^m_k $ considered as topological spaces here:
(**)
"In appropriate sense " means namely in following sense:
If we fix a field $K$ (for example $K = \mathbb{C},\mathbb{R}$) then we can consider the set of $K$ valued points $\mathbb{P}^m(K):= Hom_{Sch}(Spec(K), \mathbb{P}^m)$. And it is a fact that $\mathbb{P}^m(K)$ can be endowed with topology making $\mathbb{P}^m(K)$ homeomorphic to the "topological" projective space as defined here:
https://en.wikipedia.org/wiki/Projective_space#Definition_of_projective_space
Since $\mathbb{P}^m \to \mathbb{P}^m(K)$ is functorial we obtain from $\varphi$ as defined in (*) a new map $\tilde{\varphi}: \mathbb{P}^m(K) \to \mathbb{P}^n(K)$.
My question is why and how to see that $\tilde{\varphi}$ coinsides exactly with map given here in (**)?
For sake of simplicity assume $K =\mathbb{R}$.