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

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.

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, $$\tau_{X}\subseteq \tau$$.