To prove that $f \circ g: \mathbb{Z} \to \mathbb{Z}$ (the composition of $f$ and $g$) is also one-to-one and onto, we need to verify both properties explicitly.

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1. Proof of One-to-One (Injectivity):

Let $f: \mathbb{Z} \to \mathbb{Z}$ and $g: \mathbb{Z} \to \mathbb{Z}$ be one-to-one. We must show that $f \circ g$ is one-to-one.

Argument:

Suppose $f \circ g(x_1) = f \circ g(x_2)$ for some $x_1, x_2 \in \mathbb{Z}$. By definition of composition: $f(g(x_1)) = f(g(x_2)).$

Since $f$ is one-to-one, $f(a) = f(b)$ implies $a = b$. Thus: $g(x_1) = g(x_2).$

Similarly, since $g$ is one-to-one, $g(c) = g(d)$ implies $c = d$. Therefore: $x_1 = x_2.$

This shows that $f \circ g$ is one-to-one.

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2. Proof of Onto (Surjectivity):

Let $f: \mathbb{Z} \to \mathbb{Z}$ and $g: \mathbb{Z} \to \mathbb{Z}$ be onto. We must show that $f \circ g$ is onto.

Argument:

Let $y \in \mathbb{Z}$ be an arbitrary element of the codomain. Since $f$ is onto, there exists some $z \in \mathbb{Z}$ such that: $f(z) = y.$

Since $g$ is onto, there exists some $x \in \mathbb{Z}$ such that: $g(x) = z.$

Substituting $g(x)$ for $z$ in $f(z) = y$, we have: $f(g(x)) = y.$

This shows that for any $y \in \mathbb{Z}$, there exists an $x \in \mathbb{Z}$ such that $f \circ g(x) = y$. Hence, $f \circ g$ is onto.

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Conclusion:

Since $f \circ g$ is both one-to-one and onto, we conclude that $f \circ g: \mathbb{Z} \to \mathbb{Z}$ is a bijection.

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Would you like further clarification or examples? Here are some related questions to expand your understanding:

1. What are examples of one-to-one and onto functions $f$ and $g$?
2. How does the composition of non-injective functions behave?
3. How does the composition of non-surjective functions behave?
4. Can we generalize this result to functions between arbitrary sets?
5. How does the proof change if $f$ or $g$ are not bijections?

Tip: Remember that function composition preserves injectivity and surjectivity individually, which is why the composition of bijections is also a bijection!