Let $\mathcal{F}$ be the family of bijective increasing functions $f\colon [0,1] \to [0,1]$, and let $a \in (0,1)$. Determine the best constants $m_a$ and $M_a$, such that for all $f \in \mathcal{F}$ we have
\[m_a \leq f(a) + f^{-1}(a) \leq M_a.\]

[i](Dan Schwarz)[/i]
1. **Determine \( M_a \):**

   - For \( a = 0 \):
     \[
     f(0) = 0 \quad \forall f \in \mathcal{F}
     \]
     Therefore,
     \[
     f(0) + f^{-1}(0) = 0 + 0 = 0
     \]
     Hence, \( M_0 = 0 \).

   - For \( a = 1 \):
     \[
     f(1) = 1 \quad \forall f \in \mathcal{F}
     \]
     Therefore,
     \[
     f(1) + f^{-1}(1) = 1 + 1 = 2
     \]
     Hence, \( M_1 = 2 \).

   - For \( 0 < a < 1 \):
     - If \( f(a) = a \), then
       \[
       f(a) + f^{-1}(a) = a + a = 2a \leq a + 1
       \]
     - If \( f(a) < a \), then \( f^{-1}(a) > a \) and
       \[
       f(a) + f^{-1}(a) < a + 1
       \]
     - If \( f(a) > a \), then \( f^{-1}(a) < a \) and
       \[
       f(a) + f^{-1}(a) < a + 1
       \]
     Therefore,
     \[
     M_a \leq a + 1
     \]
     To show that \( M_a = a + 1 \) is achievable, consider the function \( f_n(x) \) defined by the points \((0,0)\), \((a - \frac{1}{n}, a)\), \((a, 1 - \frac{1}{n})\), and \((1,1)\). For large \( n \),
     \[
     f_n(a) + f_n^{-1}(a) = 1 + a - \frac{2}{n}
     \]
     As \( n \to \infty \), \( f_n(a) + f_n^{-1}(a) \to a + 1 \). Hence, \( M_a = a + 1 \).

2. **Determine \( m_a \):**

   - For \( a = 0 \):
     \[
     f(0) = 0 \quad \forall f \in \mathcal{F}
     \]
     Therefore,
     \[
     f(0) + f^{-1}(0) = 0 + 0 = 0
     \]
     Hence, \( m_0 = 0 \).

   - For \( a = 1 \):
     \[
     f(1) = 1 \quad \forall f \in \mathcal{F}
     \]
     Therefore,
     \[
     f(1) + f^{-1}(1) = 1 + 1 = 2
     \]
     Hence, \( m_1 = 2 \).

   - For \( 0 < a < 1 \):
     - If \( f(a) = a \), then
       \[
       f(a) + f^{-1}(a) = a + a = 2a \geq a
       \]
     - If \( f(a) < a \), then \( f^{-1}(a) > a \) and
       \[
       f(a) + f^{-1}(a) > a
       \]
     - If \( f(a) > a \), then
       \[
       f(a) + f^{-1}(a) > a
       \]
     Therefore,
     \[
     m_a \geq a
     \]
     To show that \( m_a = a \) is achievable, consider the function \( f_n(x) \) defined by the points \((0,0)\), \((a, \frac{1}{n})\), \((a + \frac{1}{n}, a)\), and \((1,1)\). For large \( n \),
     \[
     f_n(a) + f_n^{-1}(a) = a + \frac{2}{n}
     \]
     As \( n \to \infty \), \( f_n(a) + f_n^{-1}(a) \to a \). Hence, \( m_a = a \).

\(\blacksquare\)

The final answer is \( \boxed{ m_a = a } \) and \( M_a = a + 1 \).