I'm doing this exercise and getting confused by the definition of the initial topology. What does it mean by "the initial topology on X with respect to the maps $\mathbf{d_\tilde{x}}$"?
3 Answers
The initial topology on $X$ defined in the second box is also sometimes called the initial topology on $X$ with respect to the family $(f_j)_{j\in J}$.
Thus, the initial topology on $X$ with respect to the family $(d_\tilde{x})_{\tilde x \in X}$ is the topology that satisfies points (a) and (b) in the definition you are given, where $J = X$, and the family of functions is $(d_\tilde{x})_{\tilde x \in X}$.

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In the definition of initial topology, you used a family of functions $(f_j)_j$. The definition you gave should be that the topology on $X$ that meets those conditions (which depend on the $f_j$) should be called "the initial topology with respect to the functions $f_j$".
Applying this new definition to the exercise yields that the metric topology on the metric space $X$ should coincide with the topology for which all $d_{\overline x}$ are continuous, and for every topological space $Z$ and every function $g:Z\to X$, $g$ is continuous iff all $d_{\overline x}\circ g$ are continuous.

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In this answer I gave an extensive overview of the initial topology and its properties (your defining continuity property is called the universal mapping property for initial topologies there).
First note that all maps $d_\tilde{x}$ are indeed continuous wrt the metric topology, as they're even Lipschitz: $|d_\tilde(x)(x) - d_\tilde(x)(y)| \le d(x,y)$ for all $x,y, \tilde{x} \in X$.
And if $\mathcal{T}$ is a topology that makes all $d_\tilde(x)$ continuous, then for all $X \in X, r>0$: $$B(x,r) = d_x^{-1}[(\leftarrow,r)]\in \mathcal{T}$$ which shows that the metric topology is a subset of any such $\mathcal{T}$ and so by the minimality and unicity of the initial topology (as I defined and proved in the linked answer above), the metric topology must equal the initial topology wrt all maps $d_x$.

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