Product topology and subspace topology

Question

This seems to be a basic question but I can't get a proof.

Suppose $X\times Y$ is the product space of non-empty topological spaces $(X, \tau)$ and $(Y, \rho)$. Let $\tau'$ and $\rho'$ be the subspace topologies of $X$ and $Y$, derived from the product topology through the inclusion maps $\iota_1: X\to X\times Y$ and $\iota_2: Y\to X\times Y$, respectively. Is it always true $\tau'=\tau$ and $\rho'=\rho$?

If there are infinitely many, of arbitrary infinite cardinality, factor spaces, does the above still hold?

If the above is indeed true, can we then think of this as the defining property for product topology? That is, suppose we start from an arbitrary topology $\eta$ on the set $X\times Y$, and equip $X$ and $Y$ with subspace topologies $\eta|_X$ and $\eta|_Y$, derived from $\eta$ through the inclusion maps $\iota_1$ and $\iota_2$, respectively. Is it not necessarily true that $\eta= \eta|_X \times \eta|_Y$?

This question was motivated by this. Thanks in advance.

Answer 1

Based on the comment by @Frousse, the correct statement should be as the following.

Proposition. Let $(X_\alpha, \tau_\alpha)_{\alpha\in A}$ be a family of non-empty topological spaces with index set $A$, and $X:=\prod_{\alpha\in A}X_\alpha$ the product space with the product topology $\tau$ and projection maps $\pi_\alpha: X\to X_\alpha$. Fix $\hat{x}:=\left(\hat{x_\alpha}\right)_\alpha\in X$, where $\hat{x_{\alpha}}\in X_{\alpha}$ for all $\alpha\in A$, and define corresponding inclusion maps $\iota_{\alpha, \hat{x}}: X_\alpha \to X$ such that $\left(\pi_{\beta}\circ\iota_{\alpha, \hat{x}}\right)(x_\alpha)=\begin{cases}\hat{x_{\beta}},& \quad \beta\neq \alpha\\ x_\alpha,& \quad \alpha=\beta \end{cases}$. If equipping each image $Y_{\alpha, \hat{x}}:=\iota_{\alpha, \hat{x}}\left(X_\alpha\right)\subset X$ with their subspace topology $\tau|_{Y_{\alpha, \hat{x}}}$ respectively, then $\tau\simeq\prod_{\alpha\in A}\tau|_{Y_{\alpha, \hat{x}}}$.

The proof requires a basic lemma.

Lemma. A net (sequence) $(x_i)_{i\in I}$ converges in the product space $X=\prod_{\alpha\in A}X_\alpha$ iff the component nets $\left(\pi_{\alpha}(x_i)\right)_{i\in I}$ converges in $X_\alpha$ for all $\alpha\in A$, where $\pi_\alpha: X\to X_\alpha$ are the set-theoretic projections. (That is, product topology is the topology of component-wise convergence. A proof can be found here.)

Proof of proposition. It suffices to show every inclusion $\iota_{\alpha, \hat{x}}: X_\alpha\to Y_{\alpha, \hat{x}}$ is a homeomorphism.

Let $U_\alpha$ be an open set in $X_\alpha$, then $\pi_\alpha^{-1}(U_\alpha)$ is open in the product space $X$, and so its image $\iota_{\alpha, \hat{x}}(U_\alpha)=Y_{\alpha, \hat{x}}\cap \pi_\alpha^{-1}(U_\alpha)$ is open under the subspace topology of $Y_{\alpha, \hat{x}}$.

Conversely, let a net $(x_\alpha^i)_{i\in I}\to \xi_\alpha$ converge in $X_\alpha$, we want to show net convergence $\iota_{\alpha, \hat{x}}(x_\alpha^i)_{i\in I}\to \iota_{\alpha, \hat{x}}(\xi_\alpha)$ in $Y_{\alpha, \hat{x}}$, which is a subset of product space $X$. By the lemma above, it suffices to show net convergence $\left(\pi_{\beta}\circ\iota_{\alpha, \hat{x}}\right)(x_\alpha^i)_{i\in I}\to \left(\pi_{\beta}\circ\iota_{\alpha, \hat{x}}\right)(\xi_\alpha)$ in $X_\beta$, for every $\beta\in A$. This is trivially true by the definition of inclusion map $\iota_{\alpha, \hat{x}}$ in the proposition.