Suppose $ A,\ B $ and $ C $ are sets and $ f: A\rightarrow B $.
Suppose that $ C $ has at least two elements, and for all functions $ g $ and $ h $ from $ B $ to $ C $, if $ g\circ f = h \circ f$ then $ g = h $. Prove that $ f $ is onto.
The answer is here: http://users.metu.edu.tr/serge/courses/111-2011/textbook-math111.pdf at pg 352 and the question on pg 244. I'm a bit confused by it. Especially this part: $\forall x\in B\setminus\{b\}(h(x) = c1)$, and $h(b) = c2$. If $b$ is suppose to be arbitary how can we create this sort of piecewise function? Anyway can anyone explain away my problem? Also any alternative proof is welcome.
Best Answer
Proof by contradiction. Suppose $f$ satisfies the given condition but is not onto, then we can find $x \in B$ such that $f^{-1}(x) = \emptyset$. Now, $C$ has at least two elements so let $c_1,c_2 \in C$ with $c_1 \neq c_2$. Let $g:B \rightarrow C$ be a function satisfying $g(x) = c_1$. Next define $h: B \rightarrow C$ a function where $$h(b) = \left\{ \begin{matrix}c_2, & \text{if $b = x$} \\ g(b), & \text{if $b \neq x$} \end{matrix} \right.$$
Now clearly $g \neq h$ since $g(x) \neq h(x)$. So let's check to see if $gf = hf$. Since the image of $f$ does not contain $x$ and $x$ is the only element for which the functions $g,h$ differ we see that $gf = hf$. The argument in symbols: let $a \in A$ then $f(a) \neq x$ so $h(f(a)) = g(f(a))$ by definition of $h$, so $hf = gf$.