Let $X,Y,Z$ be sets Suppose $f:X\rightarrow Z$ and $g: X\rightarrow Y$ are two maps. Suppose $g$ is surjective. Show that there exists a unique map $h: Y\rightarrow Z$ such that $h\circ g= f$
This problem is really a problem in elementary set theory.
Since $g$ is surjective, for each $y\in Y$ there exists atleast one $x\in X$ such that $g(x)=y$. So define $h:Y\rightarrow Z$ by specifying $h(y)=f(x)$.
Now, we need to verify that the map $h$ is well defined. In other words,
If $y_1=y_2$ then $h(y_1)=h(y_2)$.
To this end, suppose $y_1=y_2$ for $y_i\in Y$. By surjectivity of $g$, there exists some $x_1,x_2\in X$ such that $g(x_1)=g(x_2)$. Since $y_1=y_2$, we may assume $x_1=x_2$. Since $f$ is well defined, $f(x_1)=f(x_2)$ and so, $h(y_1)=h(y_2)$. Clearly, $h\circ g=f$.
Uniqueness: Suppose there exists another map $h':Y\rightarrow Z$ for which $h'\circ g=f$. I must now show that $h=h'$. For $y\in Y$, there exists some $x\in X$, such that $g(x)=y$. Hence, $h(y)=f(x)=h'(g(x))=h'(y)$.
Is this correct?