Topology induced by Uniform Structure?

Question

Let $f:X \to Y$ and $g:Y \to (Z,U^{''}) $ be mappings and $ U^{'} $ be induced uniform structures on $ Y $ by $ g $ and $ U $ be induced uniform structures on $ X $ by $ f $ , $ U_{1} $ be induced uniform structures on $ X $ by $ g \circ f $. Is $ U=U_{1} $?

Answer 1

So a base for the uniformity $\mathcal{U'}$ is $$\{(g \times g)^{-1}[O]: O \in \mathcal{U}''\}\tag{0}$$

And $\mathcal{U}$ is induced by $f$, so we must assume that $Y$ then has the uniformity $\mathcal{U}'$ and $\mathcal{U}$ has as its base

$$\{(f \times f)^{-1}[O]: O \in \mathcal{U'}\}\tag{1}$$

And $\mathcal{U}_1$ has the base

$$\{((g \circ f) \times (g \circ f))^{-1}[O]: O \in \mathcal{U}''\}\tag{2}$$

And as $$((g \circ f) \times (g \circ f))^{-1}[O] = (f \times f)^{-1}\left[(g \times g)^{-1}[O]\right]$$

we see that $\mathcal{U}_1 \subseteq \mathcal{U}$ and the other inclusion is equally obvious if you think on it.

So yes, a special of a transitive law of initial structures.