Three numbers $a,b,c$ belong to $[0,\pi /2]$ interval with $\cos a = a, \sin(\cos b) = b, \cos(\sin c ) = c$. 
Sort those numbers in increasing order.
1. **Given Conditions:**
   - \( a, b, c \in \left[0, \frac{\pi}{2}\right] \)
   - \( \cos a = a \)
   - \( \sin(\cos b) = b \)
   - \( \cos(\sin c) = c \)

2. **Lemma:**
   For all \( x \in \left[0, \frac{\pi}{2}\right] \), it holds that \( x \ge \sin x \) with equality if and only if \( x = 0 \).

3. **Observation:**
   - \( c \neq 0 \) and \( b \neq \frac{\pi}{2} \).

4. **Proof that \( a > b \):**
   - Assume \( a \le b \).
   - Then \( a = \cos a \ge \cos b \).
   - Since \( \cos b > \sin(\cos b) = b \), we have \( \cos b > b \).
   - Therefore, \( a \ge \cos b > b \ge a \), which is a contradiction.
   - Hence, \( a > b \).

5. **Proof that \( c > a \):**
   - Assume \( a \ge c \).
   - Then \( a \ge c > \sin c \).
   - Since \( a = \cos a \) and \( \cos a < \cos(\sin c) = c \), we have \( \cos a < c \).
   - Therefore, \( a \ge c > \sin c \) and \( \cos a < c \le a \), which is a contradiction.
   - Hence, \( c > a \).

6. **Conclusion:**
   - From the above proofs, we have \( a > b \) and \( c > a \).
   - Therefore, the order of the numbers in increasing order is \( b < a < c \).

The final answer is \( \boxed{ b < a < c } \).