Isomorphism of Posets Let $(X,\le),(Y,\le)$ be posets. $\text{Iso}(X,Y)$ denotes the set of isotones from $X$ to $Y$. $f:X\to Y$ is an isotone if $x_1\le x_2 \implies f(x_1)\le f(x_2)$.
$(A,\le),(B\le),(C,\le)$ are posets. I guess that $\text{Iso}(A,B\times C)\cong\text{Iso}(A,B)\times\text{Iso}(A,C).$ An isomprohism of posets is a monotone bijection whose inverse is monotone. The structure being preserved on $\text{Iso}(X,Y)$ is $f\le g$ if $f(x)\le g(x)$ for all $x\in X$.
After looking at examples my isomorphism is $$\phi:\text{Iso}(A,B\times C)\to \text{Iso}(A,B)\times\text{Iso}(A,C)
$$ defined by $\phi(f(cba))=(f(ca),f(cb))$ where $f$ is an isotone starting at the elements in its input.
Is this an acceptable isomorphism of posets?
 A: $\newcommand{\iso}{\operatorname{Iso}}$Your definition of $\varphi$ doesn’t really make sense, I’m afraid. If $$\varphi:\iso(A,B\times C)\to\iso(A,B)\times\iso(A,C)\;,$$ and $f\in\iso(A,B\times C)$, then $\varphi(f)\in\iso(A,B)\times\iso(A,C)$, so $\varphi(f)$ is an ordered pair $\langle g_f,h_f\rangle$ with $g_f\in\iso(A,B)$ and $h_f\in\iso(A,C)$. Thus, to define $\varphi$, you have to specify how $g_f$ and $h_f$ are determined from $f$. This can be done, but the notation gets a little messy; it’s easier to start with the isomorphism in the other direction, from $\iso(A,B)\times\iso(A,C)$ to $\iso(A,B\times C)$, so I’ll do that instead.
Let $$\varphi:\iso(A,B)\times\iso(A,C)\to\iso(A,B\times C)$$
be defined as follows: for each $\langle f,g\rangle\in\iso(A,B)\times\iso(A,C)$, 
$$\varphi(\langle f,g\rangle):A\to B\times C:a\mapsto\langle f(a),g(a)\rangle\;.$$
I’ll be traditionally sloppy and write $\varphi(f,g)$ for $\varphi(\langle f,g\rangle)$. Clearly $\varphi(f,g)$ is some function from $A$ to $B\times C$; we must check that it is isotone. Suppose that $a_0,a_1\in A$ with $a_0\le a_1$; $f$ and $g$ are isotone, so $f(a_0)\le f(a_1)$ and $g(a_0)\le g(a_1)$, and therefore $\langle f(a_0),g(a_0)\rangle\le\langle f(a_1),g(a_1)\rangle$ in the product partial order on $B\times C$. This shows that $\varphi(f,g)$ is indeed isotone. It remains to check that $\varphi$ is an isotone bijection. Checking the isotonicity of $\varphi$ is straightforward; I’ll leave that to you. It’s also easy enough to check that $\varphi$ is injective.
To show that $\varphi$ is surjective, let $f\in\iso(A,B\times C)$ be arbitrary; we need to find $g_f\in\iso(A,B)$ and $h_f\in\iso(A,C)$ such that $\varphi(g_f,h_f)=f$. There’s really only one plausible candidate. Let $$\pi_B:B\times C\to B:\langle b,c\rangle\mapsto b$$ and $$\pi_C:B\times C:\langle b,c\rangle\mapsto c$$ be the projections from the product $B\times C$ to the factors $B$ and $C$, respectively. Then $g_f$ should be $\pi_B\circ f:A\to B$, and $h_f$ should be $\pi_C\circ f:A\to C$; to complete the proof you need only check that 
$$\begin{align*}
&\pi_B\circ f\in\iso(A,B)\;,\\
&\pi_C\circ f\in\iso(A,C)\;,\text{ and}\\
&\varphi(\pi_B\circ f,\pi_C\circ f)=f\;.
\end{align*}$$
A: Your definition of $\phi$ doesn't really make sense (to me). An isotone $f:A\to B\times C$ gives rise to a pair of functions $g_1:A\to B$ and $g_2:A\to C$ by composing with the canonical projections $B\times C \to B$ and $B\times C\to C$. It is easily verified that $g_1,g_2$ are isotone, and that the association $f\mapsto (g_1,g_2)$ gives rise to a bijection of sets $Iso(A,B\times C)\to Iso(A,B)\times Iso(A,C)$.
Remark: All of the above follows from the fact that $B\times C$, with the obvious poset structure, is the categorical product in the category $Pos$ of posets and isotone mappings. The set $Iso(A,B)$ naturally acquires a poset structure (from $B$) and is in fact that internal hom with respect to the categorical product. In short: $Pos$ is a cartesian closed category. 
