To show that $f$ is bijective given the conditions $g \circ f = \text{id}_A$ and $f \circ g = \text{id}_B$, we can analyze the implications of these equations.

1. Definitions:

   • $\text{id}_A$ is the identity function on set $A$, meaning $\text{id}_A(x) = x$ for all $x \in A$.
   • $\text{id}_B$ is the identity function on set $B$, meaning $\text{id}_B(y) = y$ for all $y \in B$.

2. Proving $f$ is Injective:

   • Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$.
   • Applying $g$ to both sides gives $g(f(x_1)) = g(f(x_2))$.
   • By the definition of composition, $g(f(x_1)) = \text{id}_A(x_1) = x_1$ and $g(f(x_2)) = \text{id}_A(x_2) = x_2$.
   • Thus, $x_1 = x_2$, proving that $f$ is injective.

3. Proving $f$ is Surjective:

   • For any $y \in B$, we need to find an $x \in A$ such that $f(x) = y$.
   • Since $f \circ g = \text{id}_B$, we have $f(g(y)) = y$.
   • This shows that for every $y \in B$, there exists an $x = g(y) \in A$ such that $f(x) = y$.
   • Thus, $f$ is surjective.

4. Conclusion about $g$:

   • Since $f$ is both injective and surjective, it is bijective.
   • The equations $g \circ f = \text{id}_A$ and $f \circ g = \text{id}_B$ indicate that $g$ is a two-sided inverse of $f$.
   • Therefore, we conclude that $g = f^{-1}$.

Summary

The conditions $g \circ f = \text{id}_A$ and $f \circ g = \text{id}_B$ show that $f$ is a bijection, and $g$ acts as its inverse, so $g = f^{-1}$.

Would you like further details or clarification on any point? Here are some related questions to consider:

1. What are the implications of $f$ being injective and surjective?
2. How do we prove that a function has an inverse?
3. What does it mean for a function to be bijective in practical applications?
4. Can you provide examples of bijective functions and their inverses?
5. How do the concepts of injectivity and surjectivity apply in other areas of mathematics?

Tip: Always check both the injective and surjective properties to confirm that a function is bijective!