$\require{AMScd}\DeclareMathOperator\Spec{Spec}$
Consider the following two commutative diagrams where $X':=X\times_{\operatorname{Spec} k, f_k} \operatorname{Spec} k$ is the pullback of $X$ along $f_k$ the absolute frobenius of $\operatorname{Spec} k$
$$\begin{CD} X' @>f>> X \\ @VVV @VVV \\ \operatorname{Spec} k @>f_k>> \operatorname{Spec}k \end{CD}$$
and
$$\begin{CD} S @>f_S>> S \\ @VVV @VVV \\ S @>f_S>> S \end{CD}$$
Prove that this diagram is commutative
$$\begin{CD} X'\times S @>(\pi_1,f_S)>> X\times S \\ @VVV @VVV \\ S @>f_S>> S \end{CD}$$ where $f_S$ the absolute frobenius of $S$ and $\pi_1$ is the first projection.
I am trying to find a solution using 2-fibre products but I don't know where to start, can we see a commutative square as an object and construct the fibered product of two squares?
Thank you for any hep.
The Categorical Answer
There's nothing complicated here.
Just check commutativity directly. The vertical maps are the projections onto the second factor. By definition of $(\pi_1,f_S)$ (which only makes sense if I assume you really mean $\pi_1\times f_S$), we have $\pi_2(\pi_1,f_S)=f_S\pi_2$, which is what you get when you go around the other way.
The Algebraic Geometry Answer
If you want an even more explicit calculation, we have the following:
Without loss of generality you may assume that $X$ and $S$ are affine, since pullbacks (and products) are computed locally.
Then if $S=\newcommand\Spec{\operatorname{Spec}}\Spec A$, $X=\Spec B$, for $A$, $B$ $k$-algebras, and if $\phi_k : k\to k$ and $\phi_A : A\to A$ are the Frobenius maps, we have that $X' = \Spec (B\otimes_{\phi} k)$, and we need to show that the following diagram commutes: $$ \require{AMScd} \begin{CD} A @>\phi_A>> A\\ @Va\mapsto a \otimes 1 VV @VVa\mapsto a \otimes 1 \otimes 1V \\ A\otimes_k B @>>a\otimes b \mapsto \phi_A(a) \otimes b\otimes 1> A\otimes_k(B\otimes_\phi k)\\ \end{CD} $$