diff --git a/docs/en/mathematics/set-theory/maps.md b/docs/en/mathematics/set-theory/maps.md
index bd49eca..0f0f5ab 100644
--- a/docs/en/mathematics/set-theory/maps.md
+++ b/docs/en/mathematics/set-theory/maps.md
@@ -14,10 +14,10 @@ For example, let $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = \sqrt{x}$ for all $
 <br>
 
 > *Proposition*: let $f: A \to B$ and $g: B \to C$ be maps, then the composition $g$ after $f$: $g \circ f = f;g$ is a map from $A$ to $C$. 
-> 
-> ??? note "*Proof*:"
-> 
->   Let $a \in A$ then $g(f(a))$ is an element in $C$ in relation $f;g$ with $a$. If $c \in C$ is an element in $C$ that is in relation $f;g$ with $a$, then there is a $b \in B$ with $afb$ and $bgc$. But then, as $f$ is a map, $b=f(a)$ and as $g$ is a map $c=g(b)$. Hence $c=g(b)=g(f(a))$ is the unique element in $C$ which is in relation $g \circ f$ with $a$. 
+
+??? note "*Proof*:"
+ 
+    Let $a \in A$ then $g(f(a))$ is an element in $C$ in relation $f;g$ with $a$. If $c \in C$ is an element in $C$ that is in relation $f;g$ with $a$, then there is a $b \in B$ with $afb$ and $bgc$. But then, as $f$ is a map, $b=f(a)$ and as $g$ is a map $c=g(b)$. Hence $c=g(b)=g(f(a))$ is the unique element in $C$ which is in relation $g \circ f$ with $a$. 
 
 <br>
 
@@ -40,9 +40,9 @@ If $B'$ is a subet of $B$ then the pre-image of $B'$, denoted by $f^{-1}(B') is
 >
 > * If $A' \subseteq A$, then $f^{-1}(f(A')) \supseteq A'$.
 > * If $B' \subseteq B$, then $f(f^{-1}(B')) \subseteq B'$.
->
-> ??? note "*Proof*:"
->
->   Let $a' \in A'$, then $f(a') \in f(A')$ and hence $a' \in f^{-1}(f(A'))$. Thus $A' \subseteq f^{-1}(f(A'))$.
->
->   Let $a \in f^{-1}(B')$, then $f(a) \in B'$. Thus $f(f^{-1}(B')) \subseteq B'$.
\ No newline at end of file
+
+??? note "*Proof*:"
+
+   Let $a' \in A'$, then $f(a') \in f(A')$ and hence $a' \in f^{-1}(f(A'))$. Thus $A' \subseteq f^{-1}(f(A'))$.
+
+   Let $a \in f^{-1}(B')$, then $f(a) \in B'$. Thus $f(f^{-1}(B')) \subseteq B'$.
\ No newline at end of file