Is the Product of two path connected space path connected? I've tried:
Let $X ,Y$ be two path connected space. Let $I =[0,1]$. Let $(a,b),(c,d) \in X \times Y$ then $f:I\rightarrow X\times Y$ defined by $f(t)=(g(t),h(t))$ where $g$ and $h$ are path joining $a$ to $c$ in $X$ and $b$ to $d$ respectively in $Y$. Then $f$ is the required path between $(a,b),(c,d)$.
Am I wrong?
 A: You are right! That is easy to see since a map $f: Z \to X \times Y$ from an arbitrary space into a product space is continuous if and only if the maps
$$p_X \circ f: Z \to X \text{ and } p_Y \circ f : Z \to Y$$ are continuous, where 
$$p_X: X \times Y \to X, (x,y) \mapsto x$$ denotes the projection (for $p_Y$ analogous). This is easy to proof, the projections are continuous and the concatenation of maps leads to a continuous map, this shows $\Longrightarrow$.
$\Longleftarrow$ can be shown by checking continuity by definition:
A map between topological spaces is continuous if and only if it pulls back open sets onto open sets.
Thus $f$ has to pull back an open set from $X \times Y$ onto an open set of $Z$. But since the $U \times V$ for open $U$ in $X$ and open $V$ in $Y$ form a basis of the topology on $X \times Y$ (producttopology), it suffices to check, that $f$ pulls back $U \times V$ onto open $H \subset Y$. This can be done the following way:
$$f^{-1}(U \times V) = f^{-1}(p_X^{-1}(U) \cap p_Y^{-1}( V)) =  f^{-1}(p_X^{-1}(U)) \cap f^{-1}(p_Y^{-1}(V))$$
$$= (p_X \circ f)^{-1}(U)\cap (p_Y \circ f)^{-1}(V)$$.
Since $p_X \circ f$ and $p_Y \circ f$ are assumed to be continuous, $(p_X \circ f)^{-1}(U)$ and $(p_Y \circ f)^{-1}(V)$ are open in Z. As $ (p_X \circ f)^{-1}(U)\cap (p_Y \circ f)^{-1}(V)$ is just the intersection of these two (especially finitely many) sets, it is open in $Z$ and thus $f$ is continuous.
Lets get back to your case for $Z = I$.As $p_X \circ f = g$ and $p_Y \circ f = h$ holds and they both are continuous by assumption, the continuity of $f$ follows.
(PS: What i've shown is precisely what alex said, i didn't want to overrun him it's just like most topology-basic-courses don't use Categories as they are too powerful to be useful and just confuse some people in basic theory.)
