Surjective Maps and right cancellation I'm working through Jacobson's Basic Algebra I, and I have a question about Exercise 3 in Section 0.2. The first part asks the reader to "Show that $ S \xrightarrow{\rm \alpha}T $ is surjective if and only if there exist no maps $ \beta_1, \beta_2 $ of $T$ into a set $U$ such that $\beta_1 \ne \beta_2 $ but $\beta_1\alpha = \beta_2\alpha$."
I'm not sure that this is correct if $U$ has only one element. For example, let $S = \{0, 1\}$, $T = \{0,1,2\}$, and $U = \{0\}$. Let $\alpha$ be defined by the graph $\{(0,0), (1,1)\}$. Then $ T \xrightarrow{\rm \beta}U$ given by the graph $\{(0,0), (1,0), (2,0)\}$ is the only map from from $T$ to $U$. Thus there are no distinct maps $\beta_1, \beta_2$ from $T$ to $U$ such that $\beta_1\alpha = \beta_2\alpha$, and yet $\alpha$ is clearly not surjective. 
I am open to the possibility that I am missing something.
Thanks!
 A: In order to show that the characterization is false, since your $\alpha$ is not surjective, you would need to show that even though $\alpha$ is not surjective, the condition is satisfied. 
So you would  need to show that there does not exist a set $U$ and a pair of maps $\beta_1,\beta_2\colon T\to U$ such that $\beta_1\circ\alpha=\beta_2\circ\alpha$ and $\beta_1\neq\beta_2$. Equivalently, that for all sets $U$ and all $\beta_1,\beta_2\colon T\to U$, if $\beta_1\circ\alpha = \beta_2\circ\alpha$, then $\beta_1=\beta_2$. Your $U$ is an example of a set that does this, but it does not establish that $\alpha$ has the desired property: you've just come up with one example that works, but the condition requires you to prove that for every choice of $U$ and every choice of $\beta_1$ and $\beta_2$, you have $\beta_1\circ\alpha=\beta_2\circ\alpha$ implies $\beta_1=\beta_2$. (That would show that $\alpha$ satisfies the condition, even though it is not surjective). 
(In short, the condition negates the existence of a set $U$ with a given property $P$, which means that the condition is actually a universal statement about all possible choices of sets $U$, since $\neg(\exists U (P(U))$ is equivalent to $\forall U (\neg P(U))$. You've come up with one example of set $U$ for which $\neg P(U)$ holds, but an example does not prove a universal statement.)
And in fact, for your choice of $\alpha$, you will not be able to do this: pick $U=\{a,b\}$. Let $\beta_1\colon T\to U$ be given by $\beta_1(0)=\beta_1(1)=\beta_1(2)=a$, and let $\beta_2\colon T\to U$ be given by $\beta_1(0)=\beta_2(1)=a$, $\beta_2(2)=b$. Then $\beta_1\circ\alpha = \beta_2\circ\alpha$, but $\beta_1\neq\beta_2$.
