How to show that wedge sum is coproduct? I have to prove that wedge sum is coproduct in category of pointed
 spaces, but I have no idea how to do it. Should I construct that unique map from definiton,or what? Any help, please.
 A: Verify the property of a copruct, yes. Let $(X,*_X), (Y, *_Y)$ two pointed spaces and $(X \vee Y, *)$ their wedge sum. We will show, that it has the universal property of a coproduct:
Let $(Z, *_Z)$ a pointed space and $f\colon (X, *_X)\to (Z, *_Z)$, $g \colon (Y, *_Y) \to (Z, *_Z)$ two morphisms. Define $(f,g) \colon (X\vee Y, *) \to (Z, *_Z)$ by
\[
  (f,g)(u) = \begin{cases} f(u) & u \in X\\
                           g(u) & u \in Y
             \end{cases}
\]
Note, that $(f,g)$ is well-defined as $f(*_X) = g(*_Y) = (f,g)(*)$. We will show that $(f,g)$ is continuous. Let $U \subseteq X \vee Y$ be open. Then $U \cap X$ and $U \cap Y$ are open in $X$ resp. $Y$, and hence 
\[ (f,g)^{-1}(U) = f^{-1}(U\cap X) \cup g^{-1}(U\cap Y) \]
is open in $Z$. Obviously $(f,g)\circ i_X = f$ and $(f,g) \circ i_Y = g$ ($i_X$ resp. $i_Y$ denoting the injections). If now $h\colon(X\vee Y,*) \to (Z,*_Z)$ is any map with $hi_X = f$, $hi_Y = g$, we have for $x \in X$
\[ h(x) = hi_X(x) = f(x) \]
and for $y \in Y$
\[ h(y) = hi_Y(y) = g(y) \]
hence, $(f,g)$ is unique and $(X\vee Y, *)$ is the coproduct.
A: Yes, given maps $f_i:\langle X_i,p_i\rangle\to\langle Y,p\rangle$, I would simply construct the map $f:\left\langle\bigvee_iX_i,p^*\right\rangle\to\langle Y,p\rangle$ such that $f_i=f\circ e_i$ for each $i$, where $e_i:\langle X_i,p_i\rangle\to\left\langle\bigvee_iX_i,p^*\right\rangle$ is the canonical embedding. This is completely straightforward, so there’s no need to look for anything fancy.
Specifically, if $p^*\ne x\in\bigvee_iX_i$, there is a unique $i$ such that $x\in X_i\setminus\{p_i\}$, and we set $f(x)=f_i(x)$. (Here I’m identifying each $X_i\setminus\{p_i\}$ with its image in the quotient $\bigvee_iX_i$, but of course you can keep the quotient map explicit if you wish.) Clearly we also need to set $f(p^*)=p$; since $f_i(p_i)=p$ for each $i$, this will cause no problems. Verifying continuity of $f$ is straightforward, and the construction ensures that $f_i=f\circ e_i$ for each $i$, so it only remains to verify that $f$ is unique, which is also very straightforward to do: you quickly discover that the condition $f_i=f\circ e_i$ for each $i$ imply the desired uniqueness.
