Characterization for Topological Evaluation Mapping to be Embedding/Sufficient Condition

Theorem
Let $X$ be a topological space.

Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.

Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.

Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$.

Let $f : X \to Y$ be the evaluation mapping defined by:
 * $\forall x \in X : \map f x = \family{\map {f_i} x}_{i \mathop \in I}$

Let:
 * $(1)\quad$ the topology on $X$ be the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
 * $(2)\quad$ the family $\family {f_i}$ separate points

Then:
 * $f$ is an embedding

Proof
From User:Leigh.Samphier/Topology/Evaluation Map is Injective iff Mappings Separate Points:
 * $f$ is an injection

From Injection to Image is Bijection:
 * $f \restriction_{X \times f \sqbrk X} \mathop : X \to f \sqbrk X$ is a bijection

From User:Leigh.Samphier/Topology/Every Topological Evaluation Mapping is Continuous:
 * $f$ is continuous

From Continuity of Composite of Inclusion on Mapping:
 * $f \restriction_{X \times f \sqbrk X}$ is continuous

Let $\SS = \set{ f_i^{-1} \sqbrk V : i \in I, V \subseteq Y_i \text{ is open}}$.

Let $f_i^{-1} \sqbrk V \in \SS$ for some $i \in I, V \subseteq Y_i$ is open.

Let $\pr_i$ denote the projection from $Y$ to $Y_i$.

We have:

By defintion of product topology:
 * $\pr_i^{-1} \sqbrk V$ is open in $Y$

By definition of subspace:
 * $f \sqbrk {f_i^{-1} \sqbrk V} = \pr_i^{-1} \sqbrk V \cap f \sqbrk X$ is open in $f \sqbrk X$

By definition of restriction:
 * $f \restriction_{X \times f \sqbrk X} \sqbrk {f_i^{-1} \sqbrk V} = f \sqbrk {f_i^{-1} \sqbrk V}$

We have shown that:
 * $\forall U \in \SS : f \restriction_{X \times f \sqbrk X} \sqbrk U \text{ is open in } Y$

By definition of initial topology:
 * $\SS$ is a sub-basis for the topology on $X$

From Injection is Open Mapping iff Image of Sub-Basis Set is Open:
 * $f \restriction_{X \times f \sqbrk X}$ is an open mapping

By definition, $f \restriction_{X \times f \sqbrk X}$ is a homeomorphism.

By definition, $f$ is an embedding.