Question

Let \(A\) and \(B\) be nonempty sets, and let \(F: A \rightarrow B\) be a function. Prove that the following are equivalent: (a) \(F\) is \(I-I\). (b) There is a function \(G: B \rightarrow A\) such that \(G \circ F=I d_{A}\). (c) For any set \(C\) and for functions \(H_{1}: C \rightarrow A\) and \(H_{2}: C \rightarrow A,\) if \(F \circ H_{1}=F \circ\) \(\mathrm{H}_{2},\) then \(\mathrm{H}_{\mathrm{l}}=\mathrm{H}_{2}\)

Short answer

All conditions are equivalent due to the related implications between injectivity and left inverse properties.

Step-by-step solution

Understand Key Concepts

We need to show equivalence between three conditions related to the function \(F:A \rightarrow B\). Specifically, (a) \(F\) is injective (one-to-one), (b) There exists a function \(G: B \rightarrow A\) such that \(G \circ F = I_d(A)\), and (c) for any set \(C\) and functions \(H_1, H_2: C \rightarrow A\), if \(F \circ H_1 = F \circ H_2\), then \(H_1 = H_2\). Understanding these conditions is crucial to forming the proof logic.

Show (a) implies (b)

Assume \(F\) is injective. We need to construct a function \(G: B \rightarrow A\) such that \(G \circ F = I_d(A)\). For each \(b \in B\) that is an image of some \(a \in A\) under \(F\), define \(G(b) = a\). Due to the injectivity of \(F\), \(G\) is well-defined. Thus, for any \(a \in A\), \(G(F(a)) = a\), showing \(G \circ F = I_d(A)\).

Show (b) implies (c)

Assume \(G: B \rightarrow A\) is such that \(G \circ F = I_d(A)\). Let functions \(H_1, H_2: C \rightarrow A\) satisfy \(F \circ H_1 = F \circ H_2\). Applying \(G\) after \(F\), we get \((G \circ F) \circ H_1 = (G \circ F) \circ H_2\) which simplifies to \(I_d(A) \circ H_1 = I_d(A) \circ H_2\). Thus, \(H_1 = H_2\), proving condition (c).

Show (c) implies (a)

Assume condition (c) holds. Take any \(a_1, a_2 \in A\) such that \(F(a_1) = F(a_2)\). Let \(C\) be a set containing a single element and define \(H_1\) and \(H_2\) to be constant functions on \(C\) with \(H_1(c) = a_1\) and \(H_2(c) = a_2\). Then \(F \circ H_1 = F \circ H_2\). By condition (c), \(H_1 = H_2\), so \(a_1 = a_2\). Hence, \(F\) is injective as required.

Key concepts

Function Composition

Function composition is an essential concept in mathematics and refers to the operation of applying one function to the results of another. When we say "composing functions," it means we take the output from one function and use it as the input for a second function. For example, if we have two functions, say \(f : A \rightarrow B\) and \(g : B \rightarrow C\), the composition of \(f\) and \(g\) is denoted as \(g \circ f : A \rightarrow C\). Here, \((g \circ f)(a) = g(f(a))\) for any \(a\) in \(A\). This creates a pathway from \(A\) to \(C\) through \(B\).

• Composition is associative, meaning \((h \circ g) \circ f = h \circ (g \circ f)\) for functions \(f, g,\) and \(h\).
• Using identity functions, composition can yield the function itself; i.e., \(f \circ I_d = f = I_d \circ f\), where \(I_d\) is the identity function.

Understanding function composition is vital to grasp more complex ideas, such as inverses and how different conditions in mathematics are interconnected.

Inverse Functions

Inverse functions provide a way to "reverse" the effect of a function. If \(f : A \rightarrow B\) is a function, an inverse function \(g : B \rightarrow A\) satisfies the condition \(g \circ f = I_d(A)\), where \(I_d(A)\) is the identity function on set \(A\). This identity function acts as a neutral element, leaving elements unchanged.To find an inverse, a function must be bijective (both injective and surjective). However, the core focus here is on injectivity, which ensures that distinct inputs map to distinct outputs.

• An injective (one-to-one) function has a well-defined inverse once it is determined which elements result from each input.
• If we know a function is injective, we can construct an inverse for its image onto its domain.

In the context of our exercise, noting that \(F\) is injective allows the proper construction of the function \(G\) such that \(G \circ F = I_d(A)\). This direct mapping shows \(G\) reverses \(F\)'s effect back to the identity on \(A\).

Equivalence of Statements

In mathematics, proving statements are equivalent means showing that they each imply the others. In the given exercise, we have three statements about a function \(F : A \rightarrow B\) that need to be proven interchangeable or equivalent.(a) \(F\) is injective.(b) There exists a function \(G : B \rightarrow A\) such that \(G \circ F = I_d(A)\).(c) For any set \(C\) and functions \(H_1, H_2 : C \rightarrow A\), if \(F \circ H_1 = F \circ H_2\) then \(H_1 = H_2\).

• The equivalence is demonstrated by showing the implications: (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a).
• Step-by-step equivalence involves logical reasoning and understanding how injectivity influences the structure of functions involved.

Understanding the equivalency between these statements helps grasp not only how a single property like injectivity can be expressed in multiple ways, but also solidifies our understanding of how different mathematical concepts are interrelated through logic.