A question on the proof of an embedding theorem (point-set topology)

Question

I'm trying to understand the proof of the following theorem:

Theorem A: For each $\alpha\in A$, let $f_\alpha\colon X\to X_\alpha$. Then the evaluation map $e\colon X\to \prod X_\alpha$ is an embedding if, and only if, $X$ has the weak topology given by the functions $f_\alpha$ and the collection $\{f_\alpha\ \vert\ \alpha\in A\}$ separates points.

However, I'm having trouble understanding one crucial step in the $(\Rightarrow)$ of the proof. Further, I know the topology on $e(X)\subseteq\prod X_\alpha$ is the weak topology induced by the restricted projections, i.e. the weak topology induced by the family of maps $\{\pi_\alpha\vert_{e(X)}\colon e(X)\to X_\alpha\}_{\alpha}$. But, how does the assumption that $e$ is an embedding (in particular a homeomorphism onto $e(X)$) let us conclude that $X$ has the weak topology induced by the family of maps $\{\pi_\alpha\circ e\colon X\to X_\alpha\}_{\alpha}$?

From this point, the ($\Rightarrow$) direction of Theorem A follows directly from the observation that $\pi_\alpha\circ e = f_\alpha$, hence, $X$ has the weak topology induced by the family of maps $\{f_\alpha\colon X\to X_\alpha\}_{\alpha}$. The fact that $\{f_\alpha\ \vert\ \alpha\in A\}$ separates points is clear to.

Answer 1

Let us begin by establishing some terminology: denote by $\tau_X$ the topology of the space $X$; $\tau_\alpha$ will be the topology of the space $X_\alpha$, and lastly $\tau$ is the weak topology induced by the family $\{f_\alpha : \alpha \in A\}$; in other words, $\tau$ is the topology in $X$ generated by the subbbasis $\mathcal{S}=\{f_\alpha^{-1}[U_\alpha] : \alpha\in A, U_\alpha\in\tau_\alpha\}$. Our goal is to suppose that $e$ is an embedding and show that $\tau_X=\tau$.

First, let's focus on verifiyng the inclusion $\tau\subseteq \tau_X$. In order to do this, it suffices to show that $\mathcal{S}\subseteq \tau_{X}$. With this in mind, fix $\alpha\in A$ and $U_\alpha\in\tau_\alpha$. Observe that $e:(X,\tau_{X}) \to \prod X_\alpha$ and $\pi_\alpha: \prod X_\alpha \to X_\alpha$ are both continuous maps. Then, the equality $\pi_\alpha \circ e = f_\alpha$ tells us that $f_\alpha^{-1}[U_\alpha]= e^{-1}[\pi_\alpha^{-1}[U_\alpha]]$ is an open subset of $(X,\tau_{X})$; equivalently, $f_\alpha^{-1}[U_\alpha]\in \tau_{X}$. This proves that $\mathcal{S}\subseteq \tau_{X}$ and so, we have $\tau\subseteq \tau_X$.

For the other inclusion, take $U\in \tau_{X}$ and $x\in U$. Since $e$ is an embedding, we know that $e[U]$ is an open subset of $e[X]$. Now take $F$, a non-empty finite subset of $A$, and for every $\alpha\in F$, $U_\alpha\in\tau_\alpha$ in such a way that $$e(x)\in \left(\bigcap_{\alpha\in F} \pi_{\alpha}^{-1}[U_\alpha]\right) \cap e[X] \subseteq e[U] \quad ...\ (*)$$

I claim that $$x\in \bigcap_{\alpha\in F} f_{\alpha}^{-1}[U_\alpha] \subseteq U.$$ To see this, simply apply $e^{-1}$ everywhere on (*) and use the identity $\pi_\alpha \circ e = f_\alpha$. This shows that for every nonempty element of $\tau_{X}$ and every point in it, you can fit in-between a finite intersection of elements in $\mathcal{S}$; in other words, $\tau_{X}\subseteq \tau$.