Hi everyone is my first time working with posets and I'd like to know if the solution of the next exercise is really correct or maybe there is some errors that I cannot see. Thank in advance.
Exercise; Let $f:X\rightarrow Y$ be a function. Suppose that $Y$ is a poset with some ordering relation $\le_Y$. Define the relation $x\le_X x' \iff f(x)<_Yf(x') \lor x=x'$. (1) Shows that the relation turns $X$ into a partially ordered set. If we know in addition that the relation $\le_Y$ makes $Y$ totally ordered (2) does this means that the relation $\le_x$ makes $X$ into a totally ordered set? If not, (3) what additional assumption needs to be made on $f$ to ensure that $\le_X$ makes $X$ totally ordered?
(1) We need to check that $(X, \le_X)$ is a poset. Clearly the relation is reflexive. We will show that the relation is anti-symmetric. For the sake of contradiction suppose that $x\not= x'$ and, $x\le_X x'$ and $x'\le_X x$, in other words we thus have $f(x)<_Yf(x')$ and $f(x')<_Yf(x)$, i.e., $f(x)<_Yf(x)$ which is a contradiction. Now we wish to show that the relation is transitive. Suppose $x\le_X x'$ and $x'\le_X x''$ [notice that if $x'=x$ there is nothing to prove] we may assume $f(x)<_Yf(x')$. Then either $x'=x''$ or $f(x')<_Yf(x'')$. For the former the result is trivial and for the latter the result follows by the transitivity of $\le_Y$ and to show that the equality $f(x)=f(x'')$ yields a contradiction.
(2) Counterexample: Let $f: \mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}; \;(i,j)\mapsto i$, and we define the natural ordering on $\mathbb{N}$. Then clearly $\mathbb{N}$ is total. Let $(i,j), (i,k) \in \mathbb{N}\times \mathbb{N}$, where $k\not=j$, then both maps to $i$, so the elements are not comparable.
(3) We claim that the additional hypothesis that $f$ is injective it will sufficient.
Suppose in addition that $f$ is an injective function and let $x,x'\in X$, notice that if $x=x'$ there is nothing to prove, for that reason we may assume that $x\not= x'$. Then $f(x)\not= f(x')$ and since $f(x),f(x')\in Y$ and $Y$ is totally ordered, then either $f(x)<_Yf(x')$ in which case $x<_X x' $ or $f(x')<_Yf(x)$ which give us the same result just with the roles of $x, x'$ interchanging. Hence $X$ is totally ordered.