Characterization for Topological Evaluation Mapping to be Embedding

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


Then:

$f$ is an embedding

if and only if

$(1)\quad$ the topology on $X$ is the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
$(2)\quad$ the family $\family {f_i}$ separates points


Proof

Necessary Condition

Let $f$ be an embedding.


$(1)$ The Topology on $X$ is the Initial Topology

Let $f \sqbrk X$ denote the image of $f$.

Let $\tau_{f \sqbrk X}$ be the subspace topology on $f \sqbrk X$.

By definition of embedding:

$f \restriction_{X \times f \sqbrk X}$ is a homeomorphism between $X$ and $f \sqbrk X$


From Subspace of Product Space has Initial Topology with respect to Restricted Projections:

$\tau_{f \sqbrk X}$ is the initial topology on $f \sqbrk X$ with respect to the mappings $\family {\pr_i \restriction_{f \sqbrk X} : f \sqbrk X \to Y_i}_{i \mathop \in I}$


Let $\tau$ be the topology on $X$.

From Homeomorphic Topology of Initial Topology is Initial Topology:

$\tau$ is the initial topology on $X$ with respect to $\family {\pr_i \restriction_{f \sqbrk X} \circ f \restriction_{X \times f \sqbrk X} : X \to Y_i}_{i \mathop \in I}$


We have:

\(\ds \pr_i \restriction_{f \sqbrk X} \circ f \restriction_{X \times f \sqbrk X}\) \(=\) \(\ds \pr_i \circ f\) Composition of Mapping with Mapping Restricted to Image
\(\ds \) \(=\) \(\ds f_i\) Composite of Evaluation Mapping and Projection

Hence:

$\tau$ is the initial topology on $X$ with respect to $\family {f_i : X \to Y_i}_{i \mathop \in I}$

$\Box$

$(2)$ The Family $\family {f_i}$ Separates Points

By definition of embedding:

$f$ is a homeomorphism between $X$ and $f \sqbrk X$

By definition of homeomorphism:

$f$ is an injection


From Evaluation Mapping is Injective iff Mappings Separate Points:

the family $\family {f_i}$ separates points.

$\Box$


Sufficient Condition

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

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:

$\SS$ 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.

$\blacksquare$


Also see


Sources

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