# Characterization for Topological Evaluation Mapping to be Embedding/Sufficient Condition

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## 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 induced by $\family{f_i}_{i \mathop \in I}$.

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 $\family {f_i}$ separates points

Then:

- $f$ is an embedding

## Proof

From Evaluation Mapping 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 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:

\(\ds f \sqbrk {f_i^{-1} \sqbrk V}\) | \(=\) | \(\ds f \sqbrk {\paren{\pr_i \circ f}^{-1} \sqbrk V}\) | Composite of Evaluation Mapping and Projection | |||||||||||

\(\ds \) | \(=\) | \(\ds f \sqbrk {f^{-1} \sqbrk {\pr_i^{-1} \sqbrk V} }\) | Preimage of Subset under Composite Mapping | |||||||||||

\(\ds \) | \(=\) | \(\ds \pr_i^{-1} \sqbrk V \cap f \sqbrk X\) | Image of Preimage under Mapping |

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:

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.

$\blacksquare$