Exercise 1.3.3 Let $X,Y$ be sets such that there is a map from X onto Y . Show that $Y\leq_c X$.

  Proof:Let this onto map from $X$ to $Y$ be $f$.For any fixed $y\in Y$,Let $y^{set}$ be

  \begin{align*}

    y^{set}:\{x\in X|f(x)=y\}

  \end{align*}

It is easy to verify that for $y_1,y_2\in Y$,$y_1\neq y_2$,

\begin{equation}\label{eq:1}

  y_1^{set}\bigcap y_2^{set}=\emptyset

\end{equation}

And

\begin{align*}

 \bigcup y^{set}_{y\in Y}=X

\end{align*}

Then by the axiom of choice,there is a choice function from $Y$ to $\bigcup y_{y\in Y}^{set}$.And according to \ref{eq:1},this choice function is one-to-one.Done.