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.