Due to the large number of results, these notes will only cover proofs I think are interesting. I am a little pedantic with the distinction between a vector space $\mathbf{V}$ and its underlying set $V$. I will use the convention that the underlying set of any vector space $\mathbf{V}$ is the non-italicized $V$, and vice versa.
A function $f : A \rightarrow B, \space B\neq \empty$ is a subset of $A \times B$ such that $\forall a \in A, \exists! b \in B : (a, b) \in f$.
Image. The image of $X\sube A$ under $f$ is defined as $f(X) := \{ f(x) : x \in X\}$.
Preimage. The preimage of $Y \sube B$ under $f$ is defined as $f^{-1}(Y) := \{a \in A : f(a) \in Y\}$.
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Definition. The restriction of $f$ to $X \sube A$ is the subset of $f$ containing only domain values in $X$.
$$ f \restriction _X :=f \space \cap \space (X \times B) $$
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Definition. The set $B^A$ denotes the set of functions $f : A \rightarrow B$.
$$ B^A := \{g \sube A \times B : \forall a \in A, \exists ! b \in B : (a, b) \in g \} $$
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Theorem. For $f : A \rightarrow B$ and nonempty sets $A$ and $B$,
$$ \text{1. $f$ is injective } \iff \exists g : B \rightarrow A : g \circ f = \text{id}_A \hspace{94pt} \\
\text{2. $f$ is surjective } \iff \exists h : B \rightarrow A : f \circ h = \text{id}_B \hspace{88pt} \\
\text{3. $f$ is bijective } \iff \exists f^{-1} : B \rightarrow A : f^{-1} \circ f = \text{id}_A \space \wedge \space f \circ f^{-1} = \text{id}_B $$
$\text{(1)}$ follows from the fact that each $b \in B$ uniquely corresponds to an $a \in A$ under an injection. $(2)$ follows from the fact that functions are defined such that all $a \in A$ have a unique image; since all $b \in B$ are mapped to under a surjection, then such an inverse $h$ is well-defined and thus exists. $(3)$ follows from the conjunction of both sides of the biconditionals.
A field $\mathbf{F} = (F, 0, 1, +, \cdot)$ contains a set $S$ along with 2 binary operations, $+ : F \times F \rightarrow F$ and $\cdot : F \times F \rightarrow F$. The structure has the following axioms.
$$ \text{F1. $\textbf{Associativity}$. } a+(b+c) = (a+b)+c \\
\text{F2. $\textbf{Commutativity}$. } a+b = b+a \hspace{39pt} \\
\hspace{44pt}\text{F3. $\textbf{Identities.}$ } \exists 0 \in F: a+0=a \space \wedge \exists 1 \in F : a \cdot 1 =a \\
\hspace{21pt}\text{F4. $\textbf{Additive Inverse}$. } \exists ! -a \in F : a+(-a)=0 \\
\hspace{68pt}\text{F5. $\textbf{Multiplicative Inverse}$. } \exists!a^{-1} \in F : a\cdot a^{-1} = 1 \hspace{10pt} a\neq0 \\
\text{F6. $\textbf{Distributivity}$. } a \cdot (b+c) = a\cdot b + a \cdot c \hspace{3pt} $$
A vector space $\mathbf{V} = (V, +, \cdot)$ over a field $\mathbf{F}$ contains a set $S$ along with 2 binary operations, $+ : V \times V \rightarrow V$ and $\cdot : F \times V \rightarrow V$. The structure has the following axioms.
$$ \text{V1. $\textbf{Associativity}$. } u + (v + w) = (u + v) + w \text{ and } \lambda(\mu v) = (\lambda \mu)v\\
\text{V2. $\textbf{Commutativity}$. } u + v = v + u \hspace{127pt} \\
\text{V3. $\textbf{Identities}$. } \exists! 0 \in V : u +0=u \wedge \exists! 1 \in F : 1v =v \hspace{50pt} \\
\text{V4. $\textbf{Additive Inverse}$. } \exists!-u \in V:u+(-u) = 0 \hspace{69pt} \\
\text{V5. $\textbf{Distributivity}$. } (\lambda + \mu)u = \lambda u + \mu u \text{ and } \lambda(u+v)=\lambda u + \lambda v \hspace{2pt} $$