Are slices $\left\{b\right\}\times F\subset B\times F$ homeomorphic to $F$? Looking at a continuous projection $B\times F\rightarrow B$, are slices $\left\{b\right\}\times F\subset B\times F$ homeomorphic to $F$?
 A: Based on Matias answer, the topology on $\{b\}\times F$ is given by the set of the $\{b\}\times U$ where $U$ is in the topology of $F$. Therefore, if you consider $$\pi: \{b\}\times F\longrightarrow F$$
defined by $$\pi((b,x))=x,$$
then, obviously, if $U$ is an open of $F$,
$$\pi(\{b\}\times U)=U\in \mathcal T_F$$
and $$\pi^{-1}(U)=\{b\}\times U\in \mathcal T_{\{b\}\times F},$$
where I denote $\mathcal T_A$ the topology of a set $A$. Since the bijectivity of $\pi$ is clear, you get the result.
A: (1). Let $T_B$ be the topology on $B$ and $T_F$ be the topology on $F.$ The Tychonoff product topology $T_{B\times F}$ on $B\times F$ is defined to be the weakest topology such that the projections $\pi_B:B\times F$ and $\pi_F:B\times F$ are continuous.
(2). So $c\times f=$  $(c\times F)\cap (B\times f)=$ $((\pi^{-1})(c\times F))\cap ((\pi_F)^{-1}(B\times f)\in$ $ T_{B\times F}$  whenever $c\in B$ and $f\in F.$ This implies $$T_{B\times F}\supset D=\{c\times f:c\in B\land f\in F\}.$$ Now $ D$ is a base for a topology $T^*$ on $B\times F,$ so  $T^*\subset T_{B\times F}.$ But $\pi_B$ and $\pi_F$ are continuous with respect to $T^*,$ so by the def'n of $T_{B\times F}$ we must also have $T_{B\times F}\subset T^*.$ Therefore $T^*=T_{B\times F}.$
(3). For any $b\in B,$ the restriction $g_b$ of the projection $\pi_F$ to the subspace $\{b\}\times F$ is a continuous (because  $\pi_F$ is continuous) and it is a bijection.
(4). In order for a continuous bijection $h:I\to J$ to be a homeomorphism, it is necessary and sufficient that it be an open mapping, that is,it maps open sets to open sets. Equivalently, that there exists a base $K_I$ for $I$ such that the image $h(L)$ is open in $J$ for every $L\in K_I.$
(5). Observe that  $D$ (see (2)) is a base for  $B\times F$ so $\{b\}\times T_F$ is a base for $\{b\}\times F.$ And for every $f\in T_F,$ the image of $\{b\}\times f$, under the mapping $g_b $ (see (3)), is $f$ , which is open in $F$ (of course). Therefore  by (3) and (4), $g_b$ is a homeomorphism.
Remark. For an infinite product $V=\prod_{a\in A}S_a$ of spaces (where $A$ is an infinite set and each $S_a$ is a space), the Tychonoff-product  topology is the weakest topology on $V$ such that each projection  $\pi_a:V\to S_a$ is continuous. This is generally (much) weaker than the box-product topology on $V.$ A base for the box-product topology on $V$ is the collection of all $\prod_{a\in A}U_a,$ where $U_a$ is open in $S_a$ for each $A.$ But the Tychonoff-product topology on an infinite product has been found to be extremely useful. 
