FORM GR1268 Question 15

Question:
Let \(S,T,\) and \(U\) be nonempty sets, and let \(f:S\rightarrow T\) and \(g:T\rightarrow U\) be functions such that the function \(g\circ f:S\rightarrow U\) is one-to-one (injective). Which of the following must be true?

(A) \(f\) is one-to-one.
(B) \(f\) is onto.
(C) \(g\) is one-to-one.
(D) \(g\) is onto.
(E) \(g\circ f\) is onto.

Answer:
(A)

Answer Key:
"One-to-one" means no two elements in the domain maps to the same element in the target.
"Onto" means every element in the target is "hit" by at least one element in the domain.

If \(f\) is not one-to-one, more than one elements in \(S\) map to the same element in \(T\) (through \(f\)), and therefore, they necessarily have to map to the same element in \(U\) (through \(g\)).
Then \(g\circ f\) cannot be one-to-one.

Below, I provide a counterexample for (B) through (E).

Let
\[\begin{align}
S &= \{1, 2\}\\
T &= \{a, b, c\}\\
U &= \{x, y, z\}.
\end{align}\]

Define \(f\) as follows:
\[\begin{align}
f(1)&=a\\
f(2)&=b.
\end{align}\]

Define \(g\) as follows:
\[\begin{align}
g(a)&=x\\
g(b)&=y\\
g(c)&=y.
\end{align}\]

Then \(g\circ f\) is defined as follows:
\[\begin{align}
g\circ f(1)&=g(f(1))=g(a)=x\\
g\circ f(2)&=g(f(2))=g(b)=y.
\end{align}\]
Thus, \(g\circ f\) is one-to-one.

\(f\) is not onto because no element in \(S\) maps to \(c\in T\).
\(g\) is not one-to-one because both \(b\) and \(c\) map to \(y\in U\).
\(g\) is not onto because no element in \(T\) maps to \(z\in U\).
\(g\circ f\) is not onto because no element in \(S\) maps to \(z\in U\).