Describe fibered product in Top Fibered product always exists in $\mathsf{Top}$?
Let $X$,$Y$,$Z\in Ob(\mathsf{Top})$. Assume that continuous maps $f\colon X\rightarrow Z$,$g\colon Y\rightarrow Z$ are given. I think fibered product $X{\times}_{Z}Y$ is $\{(x,y)\in X\times Y|\, f(x)=g(y)\}$, but I don't know how to prove it. Can anybody help me out?
 A: Let $X \times_Z Y = \{ (x,y) \in X \times Y : f(x) = g(y) \}$, and $p_X, p_Y$ the projection maps from $X \times_Z Y$ to $X,Y$.  
Already $(X \times_Z Y, p_X, p_Y)$ is the fiber product of $X$ and $Y$ over $Z$ in the category of sets.  So if $E$ is a topological space, and $\phi: E \rightarrow X \times_Z Y$ is a function, you need to check that $\phi$ is continuous if and only if $p_X \circ \phi$ and $p_Y \circ \phi$ are continuous.
A: So the pullback (aka the fiber product) is an object $P$ together with two morphisms $p_1:P\to X$ and $p_2:P\to Y$ such that $f\circ p_1=g\circ p_2$. Additionally it has to satisfy the universal property. We'll get back to that.
So your hunch is correct, put
$$P:=X\times_Z Y=\{(x,y)\ |\ f(x)=g(y)\}\subseteq X\times Y$$
and let $p_1, p_2$ be two projections (these are automatically continuous), meaning $p_1(x,y)=x$ and $p_2(x,y)=y$. Then
$$(f\circ p_1)(x,y)=f(p_1(x,y))=f(x)=g(y)=g(p_2(x,y))=(g\circ p_2)(x,y)$$
meaning $f\circ p_1 = g\circ p_2$.
Now for the universal property. Assume that $Q,q_1,q_2$ is another triple such that $q_1:Q\to X$, $q_2:Q\to Y$ and $f\circ q_1=g\circ q_2$. Define
$$\phi:Q\to X\times_Z Y$$
$$\phi(x)=(q_1(x), q_2(x))$$
Note that this is well defined because $f(q_1(x))=g(q_2(x))$. It is easy to see that $\phi$ is continuous. And the universal property follows because
$$(p_1\circ\phi)(x)=p_1(\phi(x))=p_1((q_1(x), q_2(x))=q_1(x)$$
meaning $p_1\circ\phi=q_1$. Analogously $p_2\circ\phi=q_2$.
Finally note that since $p_1, p_2$ are projections then we can reverse engineer $\phi$ from $p_1\circ\phi=q_1$ and $p_2\circ\phi=q_2$. It implies that $\phi(x)=(q_1(x),q_2(x))$. That's basically the universal property of products and projections. Therefore $\phi$ is unique.
