The Stone-Čech compactification is the biggest compactification for Tychonov spaces What shown below is a reference from "Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa

Definition
Let $X$ be a Tychonov space and 
  $$
\beta=\Delta_{f\in C(X,[0,1])}f:X\rightarrow[0,1]^{C(X,[0,1])}
$$
  the diagonal function of the collection $C(X,[0,1])$ and $\beta X$ the closure of $\beta[X]$ in $[0,1]^{C(X,[0,1])}$. So the compactification $(\beta,\beta X)$ is named Stone-Čech compactification of $X$.
Theorem
If $X$ is a Tychonov space then any Hausdorff compactification $(h,K)$ of $X$ is such that $(h,K)\preccurlyeq(\beta,\beta X)$.
Proof. From previous lemma we know that $K$ is embeddable in $[0,1]^{C(K,[0,1])}$ and so without loss of generality we can suppose that $K$ is a subspace of $[0,1]^{C(K,[0,1])}$. So we consider the function
  $$
H:[0,1]^{C(X,[0,1])}\rightarrow[0,1]^{C(K,[0,1])}
$$
  defined as
  $$
[H(\xi)](g):=\xi(g\circ h)
$$
  where $g\in C(K,[0,1])$ and $\xi\in[0,1]^{C(X,[0,1])}$.
So we observe that if $g$ is an element of $C(K,[0,1])$ and if we compose $H$ with the projection $\pi_g:[0,1]^{C(K,[0,1])}\rightarrow[0,1]$ we obtain the projection $\pi_{g\circ h}:[0,1]^{C(\beta X,[0,1])}\rightarrow[0,1]$, that is $\pi_g\circ H=\pi_{g\circ h}$, and so for the universal mapping theorem for products we can claim that $h$ is continuous.
So we consider the continuous function $p:=H|_{\beta X}:\beta X\rightarrow[0,1]^{C(K,[0,1])}$ and we prove that $p\circ\beta=h$. So we pick $x\in X$ and $g\in C(K,[0,1])$ and we observe that
  $$
[(p\circ\beta)(x)](g)=[p(\beta(x))](g)=[H(\beta(x))](g)=[\beta(x)](g\circ h)=(g\circ h)(x)=[h(x)](g)
$$
  and so for the arbitrariness of $g$ we can claim that $[p\circ\beta](x)=h(x)$ for any $x\in X$, that is $p\circ\beta=h$. Then finally we observe that $p[\beta X]=K$ and so we can claim that $(h,K)\preccurlyeq(\beta,\beta X)$.

Here for the sake of completeness the original text of the proof: I hope that I made a good translation.

Well I don't understand why $p[\beta X]=K$ and why $(g\circ h)(x)=h(x)(g)$. Then in addition I don't understand why  the domain of $\pi_{g\circ h}$ is $[0,1]^{C(\beta X,[0,1])}$ rather than $[0,1]^{C(X,[0,1])}$, but this seems to be irrelevant for the proof. 
So could someone help me, please?
 A: $p \circ \beta = h$ tells you also that $p[\beta[X]] = h[X]$ and $h[X]$ is dense in $K$ (from $(h,K)$ being a compactification). So $p[\beta X]$ is a compact (hence closed in the Hausdorff $K$!) subset of $K$ that contains the dense set $h[X]$ and so must equal $\overline{h[X]}=K$. 
Note the subtlety in the proof of $p \circ \beta=h$, where we identify a point of $K$ by its "vector of values" in $[0,1]^{C(K,[0,1])}$, and that's why we work with the values at all $g \in C(K, [0,1])$.
Indeed in this context $\pi_{g \circ h}$ has domain $[0,1]^{C(X, [0,1])}$ as this is the domain on which $H$ is defined, and fits with the given diagram.
It's probably just a typo, but it "happens" that $C(X,[0,1]$ "sort of equals" $C(\beta X,[0,1])$ anyway (every $f$ defined on $X$ has a natural "extension" $\beta f$ defined on $\beta X$ (namely $\pi_f \restriction_{\beta[X]}$) and we can "restrict" any continuous map $g$ from $\beta X$ to $[0,1]$ by $g \circ \beta: X \to [0,1]$ and these operations are each other's inverses. So we have a antural isomorphism between $C(X,[0,1])$ and $C(\beta X, [0,1])$, which will in fact turn out to characterise $\beta X$ among all compactifications of $X$.
