Answer
See explanation
Work Step by Step
We know that \(g\colon Y\to Z\) is **one‐to‐one** (injective). By definition, if \(g(y_1) = g(y_2)\) for any \(y_1,y_2\in Y\), then \(y_1=y_2\). Apply this to each situation:
---
### (a) \(s_k\) and \(s_m\) are elements of \(Y\) and \(g(s_k)=g(s_m)\).
Since both \(s_k\) and \(s_m\) lie in \(Y\), injectivity directly implies
\[
s_k = s_m.
\]
---
### (b) \(\tfrac{z}{2}\) and \(\tfrac{t}{2}\) are elements of \(Y\) and \(g\!\bigl(\tfrac{z}{2}\bigr)=g\!\bigl(\tfrac{t}{2}\bigr)\).
Because \(\tfrac{z}{2},\tfrac{t}{2}\in Y\), we can apply injectivity:
\[
\frac{z}{2} = \frac{t}{2}
\quad\Longrightarrow\quad
z = t.
\]
---
### (c) \(f(x_1)\) and \(f(x_2)\) are elements of \(Y\) and \(g\bigl(f(x_1)\bigr)=g\bigl(f(x_2)\bigr)\).
Here, the two inputs to \(g\) are \(f(x_1)\) and \(f(x_2)\), both of which lie in \(Y\). Injectivity yields
\[
f(x_1) = f(x_2).
\]
Without additional assumptions about \(f\) itself, we cannot conclude \(x_1=x_2\); we only know that their images under \(f\) coincide.
