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!