diff --git a/docs/en/mathematics/set-theory/maps.md b/docs/en/mathematics/set-theory/maps.md
index 66ba566..6197662 100644
--- a/docs/en/mathematics/set-theory/maps.md
+++ b/docs/en/mathematics/set-theory/maps.md
@@ -32,7 +32,7 @@ Notice that $b$ can have more than one pre-image. Indeed if $f: \mathbb{R} \to \
 
 If $A'$ is a subset of $A$ then the image of $A'$ under $f$ is the set $f(A') = \{f(a) \;|\; a \in A'\}$, so $\text{Im}(f) = f(A)$. 
 
-If $B'$ is a subet of $B$ then the pre-image of $B'$, denoted by $f^{-1}(B') is the set of elements $a$ from $A$ that are mapped to an element $b$ of $B'$.
+If $B'$ is a subet of $B$ then the pre-image of $B'$, denoted by $f^{-1}(B')$ is the set of elements $a$ from $A$ that are mapped to an element $b$ of $B'$.
 
 <br>
 
@@ -69,3 +69,41 @@ For example the map $\sin: \mathbb{R} \to \mathbb{R}$ is not surjective nor inje
 
     Think of pigeonholes. (Not really a proof).
 
+<br>
+
+> *Proposition*: let $f: A \to B$ be a bijection. Then for all $a \in A$ and $b \in B$ we have $f^{-1}(f(a)) = a$ and $f(f^{-1}(b)) = b$. In particular, $f^{-1}$ is the inverse of $f$.
+
+??? note "*Proof*:"
+
+    Let $a \in A$. Then $f^{-1}(f(a)) = a$ by definition of $f^{-1}$. If $b \in B$ then by surjectivity of $f$ there is an $a \in A$ with $b = f(a)$. So, by the above $f(f^{-1}(b)) = f(f^{-1}(f(a))) = f(a) = b$. 
+
+<br>
+
+> *Theorem*: let $f: A \to B$ and $g: B \to C$ be two maps.
+>
+> 1. If $f$ and $g$ are surjective then so is $g \circ f$.
+> 2. If $f$ and $g$ are injective then so is $g \circ f$.
+> 3. If $f$ and $g$ are bijective then so is $g \circ f$.
+
+??? note "*Proof*:"
+
+    1. Suppose $f$ and $g$ are surjective, let $c \in C$. By surjectivity of $g$ there is a $b \in B$ with $g(b) = c$. Since $f$ is surjective there is also an $a \in A$ with $f(a) = b$. Therefore $g \circ f(a) = g(f(a)) = g(b) = c$.
+    2. Suppose $f$ and $g$ are injective, let $a,a' \in A$ with $g \circ f(a) = g \circ f(a')$. Then $g(f(a)) = g(f(a'))$ and by injectivity of $g$ we find $f(a) = f(a')$. Injectivity of $f$ implies $a = a'$.
+    3. Proofs 1. and 2. imply 3. by definition of bijectivity.
+
+<br>
+
+> *Proposition*: if $f: A \to B$ and $g: B \to A$ are maps with $f \circ g = I_B$ and $g \circ f = I_A$, where $I_A$ and $I_B$ denote the identity maps on $A$ and $B$, respectively. Then $f$ and $g$ are bijections and $f^{-1} = g$ and $g^{-1} = f$.
+
+??? note "*Proof*:"
+
+    Suppose $f A \to B$ and $g: B \to A$ are maps with $f \circ g = I_B$ and $g \circ f = I_A$. Let $b \in B$ then $f(g(b)) = b$, thus $f$ is surjective. If $a,a' \in A4 with $f(a) = f(a')$, then $a = g(f(a)) = g(f(a')) = a' and hence $f$ is injective. Therefore $f$ is bijective and by symmetry $g$ is also bijective. 
+
+<br>
+
+> *Proposition*: suppose $f: A \to B$ and $g: B \to C$ are bijective maps. Then the inverse of the map $g \circ f$ equals $f^{-1} \circ g^{-1}$.
+
+??? note "*Proof*:"
+
+    Suppose $f: A \to B$ and $g: B \to C$ are bijective maps. Then for all $a \in A$ we have $(f^{-1} \circ g^{-1}) (g \circ f)(a) = f^{-1}(g^{-1}(g(f(a)))) = f^{-1}(f(a)) = a$.
+