I constructed a following theorem and I wish to (dis)prove it:
Let $\leq$ be a total order on $A$ and let there be surjection $\varphi:A\rightarrow B$. Then there exists total order on $B$.
This seems really trivial and obvious to me. For any $b_1,b_2\in B$, because $\varphi$ is surjection, we can find $a_1,a_2\in A$ such that $\varphi(a_1)=b_1$ and $\varphi(a_2)=b_2$. Then, if I define $R\subseteq B\times B$ as follows:$$(b_1,b_2)\in R\Leftrightarrow (a_1,a_2)\in\leq$$
where, as mentioned, $b_1=\varphi(a_1),b_2=\varphi(a_2)$, am I done?
EDIT: this is obviously wrong. What I'm dealing with is a set $A$ of all possible strings of finite length of given alphabet $\Sigma$ (with some additional, not so important property, thus in fact $A\subsetneq \Sigma^*$). The total order of $\Sigma^*$ is given
1.) by the length of the strings,
2.) lexicographically (I assume, there exists some total ordering of an alphabet)
Now because $A\subsetneq \Sigma^*$ by the property of linear order, $A$ is lineary ordered too.
Now to any $b\in B$ i assign some $a\in A$ such that $a\mapsto b$. Giving surjection $A\rightarrow B$
Best Answer
If you accept the axiom of choice, then any set $B$ can be well-ordered. This gives you a total order on $B$. However, there is no relation to the order $\le_A$ on $A$, and that is probably not what you want.
Using the axiom of choice again, you get a section for $p$ which is a function $i : B \to A$ such that $p \circ i = id_B$. Obviuosly $i$ is injective. In concrete situations you may even be able to construct explicitly a section for $p$.
Given any section, the subset $i(B)$ inherits an order $\le_{i(B)}$ from $A$. This is again a total order. Now you can give $B$ the unique order $\le_i$ defined by $b \le_i b_2$ if $i(b_1) \le_A i(b_2)$. Then $(B,\le_i)$ and $(i(B),\le_{i(B)})$ are isomorphic as ordered sets. In general $\le_i$ depends on the choice of a section, so you get a whole collection of total orders which are somehow related to $\le_A$. But you cannot expect that there exists a section $i$ such that $a_1 \le_A a_2$ implies $p(a_1) \le_i p(a_2)$ for all $a_1,a_2$.
Moreover, if there exists an order $\le_B$ on $B$ (which is not required to be a total order) such that $p$ is order-preserving, then for any section $i$ the orders $\le_B$ and $\le_i$ agree. In particular, $\le_B$ is a total order.
(1) Let $b_1 \le_i b_2$. Then $i(b_1) \le_A i(b_2)$ and $b_1 = p(i(b_1) \le_B p(i(b_2) = b_2$.
(2) Let $b_1 \le_B b_2$. If $b_1 \le_i b_2$, we are done. If $b_2 \le_i b_1$, use (1) to see $b_2 \le_B b_1$. Therefore $b_1 = b_2$ and in particular $b_1 \le_i b_2$.