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Let $((X_i, \mathcal{O}_i))_{i\in I}$ be a family of topological spaces, $X$ a set and $(f_i:X\longrightarrow X_i)_{i\in I}$ a family of maps. The initial topology determined by the family $(f_i)_{i\in I}$ is the coarsest topology on $X$ which makes every map $f_i:X\longrightarrow X_i$ continuous.

I'd like to show there is indeed such a topology and to show $\{f_i^{-1}(U_i): U_i\in\mathcal{O}_i\}$ is a sub-basis for it.

For this let us call $\mathcal{O}_{0}$ the initial topology. I'm trying the following strategy:

$(i)$ First I would show the family $\mathcal{B}:=\left\{\bigcap_{j=1}^n f_{i_j}^{-1}(U_{i_j}): U_{i_j}\in\mathcal{O}_{i_j}\right\}$ induces a topology $\mathcal{O}$ in $X$ have $\mathcal{B}$ as a basis;

$(ii)$ Then I would show $\mathcal{O}=\mathcal{O}_{0}$.

For $(i)$ it suffices showing:

$(a)$ $\displaystyle X=\bigcup_{B\in\mathcal{B}} B$;

$(b)$ If $U, V\in\mathcal{B}$ and $x\in U\cap V$ then there is $W\in\mathcal{B}$ such that $x\in W\subset U\cap V$.

Then a theorem assures there exists a topology $\mathcal{O}$ on $X$ having $\mathcal{B}$ as basis (it suffices defining $\mathcal{O}$ as the family consisting of unions of elements of $\mathcal{B}$).

Can anyone help me showing $(a)$ and $(b)$?

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    $\begingroup$ Haven’t you already proved at some point that any family of subsets of $X$ is a subbase for a topology, and that the family of finite intersections of members of the subbase is a base for that same topology? (In this context $\bigcap\varnothing=X$.) Your (a) and (b) are just special cases of this basic fact. $\endgroup$ – Brian M. Scott Oct 12 '15 at 15:23
  • $\begingroup$ Well, I guess I must have seen this someday but I had forgotten. Thanks for reminding me =) $\endgroup$ – PtF Oct 12 '15 at 15:54

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