Update solution for chapter I exercise 2.5 using right-cancellation

[Copilot]

Updated the solution to chapter I, exercise 2.5 to use the correct right-cancellation definition of epimorphism.

Changes Made

• Definition: Changed from an incorrect "lifting property" style definition to the proper right-cancellation definition: A function f: A→B is an epimorphism iff for all sets Z and all functions β', β'': B→Z, if β'∘f = β''∘f, then β' = β''.

• Proof (⟹): Rewrote the proof that surjection implies epimorphism to use the right-cancellation property, showing that if f is surjective, then for any b∈B we can find a∈A with f(a)=b and prove β'(b) = β''(b).

• Proof (⟸): Rewrote the proof that epimorphism implies surjection by constructing two specific functions β', β'': B→{0,1} that distinguish elements in the image from those outside, using the cancellation property to show all elements of B must be in the image.

The solution now properly uses the right-cancellation definition, making it the categorical dual of the monomorphism (left-cancellation) definition as expected for this exercise.

Original prompt

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This section details on the original issue you should resolve

<issue_title>exercise 2.5</issue_title>
<issue_description>Won't this exercise expect a right-cancellation definition?

​</issue_description>

<agent_instructions>please modify the solution to chapter I, exercise 2.5. you should use right-cancellation definition of the epimorphism.</agent_instructions>

Comments on the Issue (you are @copilot in this section)

@hooyuser Good catch! Thanks for pointing that out—I'll fix it shortly.

• Fixes exercise 2.5 #23

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[hooyuser]

@copilot ?

[Copilot]

@copilot ?

I've updated exercise 2.5 to use the correct right-cancellation definition of epimorphism. The definition now states that f: A→B is an epimorphism iff for all sets Z and all functions β', β'': B→Z, if β'∘f = β''∘f, then β' = β''. I also rewrote both directions of the proof accordingly. See commit 86ebc1e.