User:Anghel/Sandbox

Theorem
Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.

Let $\phi: T_1 \to T_2$ be a homeomorphism.

Let $S \subseteq T_1$ be a subset of $T_1$.

Let $S$ be simply connected in $\struct {T_1, \tau_1}$.

Then $\phi \sqbrk S$ is simply connected in $\struct {T_2, \tau_2}$.

That is, simple connectedness is a topological property.

Proof
By, $S$ is path-connected.

Path-Connectedness is Preserved under Homeomorphism shows that $\phi \sqbrk S$ is path-connected.

Let $x \in S$.

Let:


 * $\gamma_1 : \closedint 0 1 \to \phi \sqbrk S$


 * $\gamma_2 : \closedint 0 1 \to \phi \sqbrk S$

be two loops in $\phi \sqbrk S$.

Let $\gamma_1, \gamma_2$ both have base point $x$.

Composite of Continuous Mappings is Continuous shows that $\phi^{-1} \circ \gamma_1$ and $\phi^{-1} \circ \gamma_2$ are continuous mappings.

So $\phi^{-1} \circ \gamma_1, \phi^{-1} \circ \gamma_2$ are loops in $S$ with base point $\map { \phi^{-1} } x $.

By, there exists a path homotopy $H: \closedint 0 1 \times \closedint 0 1 \to S$ such that:


 * $\forall s \in \closedint 0 1 : \map H { s, 0 } = \map { \phi^{-1} \circ \gamma_1 }{ s }$


 * $\forall s \in \closedint 0 1 : \map H { s, 1 } = \map { \phi^{-1} \circ \gamma_2 }{ s }$


 * $\forall t \in \closedint 0 1 : \map H { 0, t } = \map { \phi^{-1} \circ \gamma_1 }{ 0 } = \map { \phi^{-1} } x$


 * $\forall t \in \closedint 0 1 : \map H { 1, t } = \map { \phi^{-1} \circ \gamma_1 }{ 1 } = \map { \phi^{-1} } x$

Composite of Continuous Mappings is Continuous shows that $ \phi \circ H : \closedint 0 1 \times \closedint 0 1 \to \phi \sqbrk S$ is a continuous mapping.

Then:


 * $\forall s \in \closedint 0 1 : \map { \phi \circ H } { s, 0 } = \map { \phi \circ \phi^{-1} \circ \gamma_1 }{ s } = \map { \gamma_1 } s$


 * $\forall s \in \closedint 0 1 : \map { \phi \circ H } { s, 1 } = \map { \phi \circ \phi^{-1} \circ \gamma_2 }{ s } = \map { \gamma_2 } s$


 * $\forall t \in \closedint 0 1 : \map{ \phi \circ H }{ 0, t } = \map{ \phi \circ \phi^{-1} \circ \gamma_1 }{ 0 } = x$


 * $\forall t \in \closedint 0 1 : \map{ \phi \circ H }{ 1, t } = \map{ \phi \circ \phi^{-1} \circ \gamma_1 }{ 1 } = x$

It follows that $\phi \circ H$ is a path homotopy between $\gamma_1$ and $\gamma_2$.

By, $\phi \sqbrk S$ is simply connected.