Is the subspace topology of the product topology the initial topology with respect to the restrictions of the projection maps?

Question

Let $I$ be an index set, $X_i$ be a topological space for each $i \in I$ and $X = \prod_{i \in I} X_i$ the product of all $X_i$. Then the product topology is exactly the initial topology with respect to the canonical projections $\pi_i: X \to X_i$.

Let $A$ be a topological subspace of $X$ (i.e. $A$ has the subspace topology in $X$) and $\varphi_i$ are the restriciton of the projections $\pi_i$ on $A$.
Question: Is $A$ the initial topology with respect to the maps $\varphi_i$?

I think that the arguments which I need can be found in this nice explanation of Henno Brandsma, but I am unable to convert it into a line of reasoning which convinces me. Could you please answer my question and give an explanation? Thank you!

Answer 1

It's a special case of the transitive law of initial topologies, which is shown in my linked answer:

$A$ has the initial topology wrt $i_A: A \to X$.

$X$ has the initial topology wrt the maps $\pi_i: X \to X_i, i \in I$.

It follows that $A$ has the initial topology wrt the maps $\pi_i \circ i_A: A \to X_i, i \in I$, and $\pi_i \circ i_A$ is exactly $\varphi_i=\pi_i|_A$.

In that answer the original question was whether on $A=\prod_i A_i$ the product topology of subspace topologies coincides with the subspace topology of the large product, but the same law answers both questions.