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{\bf section 18}: 13) Suppose $A \subset X$.  Let $f:  A \rightarrow Y$ be 
continuous where 
$Y$ 
is $T_2$.  If 
$g_i: \overline{A} \rightarrow Y$ is continuous for $i = 1, 2$ and if
$g_1|_A = g_2|_A = f$, then $g_1 = g_2$


Pf: Suppose $g_1 \not= g_2$.  Then $\exists x \in \overline{A} - A$ such that 
$g_1(x) \not= g_2(x)$.  $Y$ $T_2$ implies $\exists V_1, V_2$ open in $Y$ 
such that $g_i(x) \in V_i$ and $V_1 \cap V_2 = \emptyset$

$x \in g_i^{-1}(V_i)$ implies $x \in  g_1^{-1}(V_1) \cap  
g_2^{-1}(V_2)$.  $g_i$ continuous implies $g_i^{-1}(V_i)$ open in 
$\overline{A}$ and hence $g_1^{-1}(V_1) \cap
g_2^{-1}(V_2)$ open in
$\overline{A}$.  Thus there exists $U$ open in $X$ such that 
$g_1^{-1}(V_1) \cap g_2^{-1}(V_2) = U \cap \overline{A}$.  Note $x \in U$.
 
$x \in \overline{A} - A$ implies $x \in A'$.  Thus $\exists a \in U \cap A - 
\{x\} \subset g_1^{-1}(V_1) \cap
g_2^{-1}(V_2)$.  But $f(a) = g_i(a) \in  g_i(g_i^{-1}(V_i)) \subset V_i$. 
But $V_1 \cap V_2 = \emptyset$.


{\bf NOTE:  It may have been simpler to WLOG assume $X = \overline{A}$, 
but 
you must state why this WLOG works. }


p. 145: 1.)  Define $*: G \times G \rightarrow G$, by $*(g_1, g_2) = g_1 
\cdot g_2$ where $\cdot$ is the group operation on $G$.

Define $i: G \rightarrow G$, by $i(g) = g^{-1}$ 

Define $id: G \rightarrow G$, by $i(g) = g$ 

Define $f:  G \times G \rightarrow G$, by $f(g_1, g_2) = g_1
\cdot g_2^{-1}$ 



Define $id: G \rightarrow G$, by $id(g) = g$ 


Define $p:  G \times G \rightarrow G \times G$, by $p(g_1, g_2) = (g_1, 
g_1^{-1})$.  





Suppose $*, i$ are continuous functions.  



Suppose $f$ is continuous.  Note ${\{e\} \times G $ (with the product 
topology) is homeomorphic to $G$.  $f|_{{\{e\} \times G}$ is 
continuous.  Note $f(e, g) = g^{-1}$.


Then 

\end
