Let us address the two parts of this problem systematically. We analyze the properties of injectivity and surjectivity for the function $f: [0, 2] \to [5, 6]$, given the conditions.

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(a) If the restrictions of $f$ to $[0, 1]$ and $[1, 2]$ are injective, must $f$ be an injection?

An injective function (or one-to-one function) ensures that every element in the domain maps to a unique element in the codomain. In mathematical terms: $f(x_1) = f(x_2) \implies x_1 = x_2.$

• Given Information:

  • $f$ is injective on $[0, 1]$, meaning no two points in $[0, 1]$ map to the same value.
  • $f$ is injective on $[1, 2]$, meaning no two points in $[1, 2]$ map to the same value.

• Analysis:

  • Injectivity on each subinterval does not guarantee that the same output value cannot occur in both $[0, 1]$ and $[1, 2]$. That is, there might exist $x_1 \in [0, 1]$ and $x_2 \in [1, 2]$ such that $f(x_1) = f(x_2)$.
  • If such a case occurs, $f$ is no longer injective on the entire domain $[0, 2]$, even though it is injective on each subinterval.

• Counterexample: Define $f(x)$ as: [ f(x) = \begin{cases} 5.5 - 0.5x, & \text{if } x \in [0, 1], \ 5.5 + 0.5(x - 1), & \text{if } x \in [1, 2]. \end{cases} ] Here, $f(x)$ is injective on $[0, 1]$ and $[1, 2]$, but $f(0) = f(2) = 5.5$, which means $f$ is not injective on $[0, 2]$.

• Conclusion: The function $f$ is not necessarily injective.

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(b) If the restrictions of $f$ to $[0, 1]$ and $[1, 2]$ are surjective, must $f$ be a surjection?

A surjective function (or onto function) ensures that every element in the codomain has at least one corresponding element in the domain. In mathematical terms: [ \forall y \in [5, 6], \exists x \in [0, 2] \text{ such that } f(x) = y. ]

• Given Information:

  • $f$ is surjective on $[0, 1]$, meaning every value in $[5, 6]$ can be attained using $x \in [0, 1]$.
  • $f$ is surjective on $[1, 2]$, meaning every value in $[5, 6]$ can be attained using $x \in [1, 2]$.

• Analysis:

  • Surjectivity on both subintervals ensures that every $y \in [5, 6]$ has at least one preimage in $[0, 1]$ and at least one preimage in $[1, 2]$.
  • Thus, $f$ is surjective on the entire domain $[0, 2]$, because $[0, 1] \cup [1, 2] = [0, 2]$.

• Counterexample Possibility: There is no way to construct a counterexample where $f$ fails to be surjective, because the union of the surjective mappings on $[0, 1]$ and $[1, 2]$ covers the entire codomain $[5, 6]$.

• Conclusion: The function $f$ must be surjective.

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Final Summary:

(a) $f$ is not necessarily injective.
(b) $f$ must be surjective.

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Do you want a detailed visualization or examples?
Here are 5 related questions to explore:

1. What conditions ensure injectivity on a domain partitioned into subintervals?
2. Can you define a function that is bijective when its restrictions are bijections?
3. How does the codomain affect the injectivity or surjectivity of a function?
4. What are practical applications of injective and surjective functions in real-world problems?
5. How does bijection relate to cardinality?

Tip: Always analyze how subinterval restrictions influence the global behavior of a function when combining properties like injectivity and surjectivity.