Question needed for clarification of notation in definition and specific term in an exercise from Arbib and Manes' basic category theory text

Question

The following question is taken from $\textit{Arrows, Structures and Functors the categorical imperative}$ by Arbib and Manes

$\textbf{9 Definition:}$ For fixed $B$. consider the class $\textbf{M}(B)$ of all $(A,f)$ with $f:A\rightarrow B$ a monomorphism. Say that $(A,f)\sim (A',f')$ if there exits $t:A\rightarrow A'$ with $f'\cdot t=f$ and $u$ with $f\cdot u=f'$. It is obvious that $\sim$ is an equivalence relation on $\textbf{M}(B)$. An equivalence class $[A,f]$ is called a $\textbf{mono-subobject}$ of $B.$

$\textbf{11 Proposition:}$ On the category $\textbf{Vect}$ (i) $f$ is an epimorphism iff $f$ is onto iff $f$ is a coequalizer. (ii) $f$ is an monomorphism iff $f$ is one-to-one iff $f$ is a equalizer.

$\textbf{Exercise: }$ Using $\textbf{11, }$ prove that for $B$ in $\textbf{Vect}$, mono-sub-objects of $B$ may be identified with subspaces of $B$. [Hint: Start by completing the proof (following 9) of the corresponding result in $\textbf{Set}$]

$\textbf{Questions: }$ I just have three quick questions.

$\textbf{(1)}$ in the definition above, The notation: $(A,f)$, does it mean the set of all maps $f:A\rightarrow B$?

$\textbf{(2)}$ The set of equivalence classes $[A,f]$ on $\sim$ in set builder notation is $[A,f]=\{(A',f')|(A',f')\sim (A,f)\}$ and the quotient set in set builder notation is written as $\textbf{M}(B)/\sim=\{[A,f]|(A,f)\in \textbf{M}(B)\}.$ But to write out explictly each each set of equivalence classes in set builder notation, do I write it like this: $[A,f]=\{(t:A\rightarrow A', u:A\rightarrow A')|f'\cdot t=f \wedge f\cdot u=f'\}.$

$\textbf{(3)}$ What does it mean to say that a "mono-sub-objects of $B$ may be $\textbf{identified}$ with subspaces of $B$?"

Thank you in advance

Answer 1

1. No. $\mathbf{M}(B)$ is the set of all monomorphisms with codomain $B$. The pair $(A,f)$ is a way of identifying both the monomorphism $f$ and its domain $A$ (you already know the codomain is $B$, because $(A,f)$ is an element of $\mathbf{M}(B)$.

2. Ugh. No. $[A,f]$ is a set whose elements are ordered pairs of the form $(A',f')$ satisfying certain conditions. What you write is a set whose elements are ordered pairs of the form $(t,u)$ where $t$ and $u$ are both maps... which is not at all what you have. $$[A,f]=\{(A',f')\mid (f'\colon A'\to B\text{ is a monomorphism})\wedge(\exists t\colon A\to A'(f'u = f))\}.$$

3. It means that every mono-sub-object of $B$ corresponds to a subspace, every subspace is a mono-sub-object, and these identifications are inverses of each other.