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> [!definition] Definition. ([[change of basepoint isomorphism]])
> Let $X$ be a [[topological space]]. Let $\alpha$ be a [[parameterized curve]] in $X$ from $x_{0}$ to $x_{1}$. The map $\begin{align}
\hat{\alpha}: \pi_{1}(X,x_{0}) & \to \pi_{1}(X,x_{1}) \\
[f] & \mapsto [\overline{\alpha}] * [f] * [\alpha]
\end{align}$
defines [[group isomorphism]] between $\pi_{1}(X,x_{0})$ and $\pi_{1}(X,x_{1})$.
\
Here, $\overline{\alpha}$ denotes the [[parameterized curve|reverse path]] of $\alpha$, $*$ is the [[fundamental groupoid|concatenation product]] operation, and $[\cdot]$ denotes the [[path homotopy]] [[equivalence class|class]] of $\cdot$.
> [!proposition] Corollary.
> If $X$ is [[path-connected]] and $x_{0}, x_{1} \in X$, then the [[fundamental group|fundamental groups]] $\pi_{1}(X,x_{0})$ and $\pi_{1}(X,x_{1})$ are [[group isomorphism|isomorphic]]. *But they are not 'naturally' so— a different path $\alpha'$ could yield a different isomorphism than $\alpha$ did above.*
> [!justification]
>
>
$\hat{\alpha}$ is well-defined because $*$ is. If $f$ is a [[parameterized curve|loop]] based at $x_{0}$, then $\overline{\alpha} * (f * \alpha)$ is a loop based at $x_{1}$, hence $\hat{\alpha}$ maps $\pi_{1}(X,x_{0})$ into $\pi_{1}(X,x_{1})$, as desired. See picture [[TODO]].
>
It is easy to verify $\hat{\alpha}$ is a [[group homomorphism]]: if $f,g$ are two [[parameterized curve|loops]] about $x_{0}$, then liberal application of associativity yields $\begin{align}
\hat{\alpha}([f] * [g]) = & \hat{\alpha}([f * g]) \\
= & [\overline{\alpha}] * [f * g] * [\alpha] \\
= & [\overline{\alpha} * f] * [g * \alpha] \\
= & [\overline{\alpha} * f] * [e_{x_{0}}] * [g * \alpha] \\
= & [\overline{\alpha} * f] * [\alpha * \overline{\alpha}] * [g * \alpha] \\
= & [\overline{\alpha} * f * \alpha] * [\overline{\alpha} * g * \alpha] \\
= & ([\overline{\alpha}] * [f] * [\alpha] ) * ([\overline{\alpha}] * [g] * [\alpha]) \\
= & \hat{\alpha}([f]) * \hat{\alpha}([g]).
\end{align}$
To show that $\hat{\alpha}$ is an [[group isomorphism|isomorphism]], we show it has inverse $\hat{\beta}$, where $\hat{\beta}$ is defined $\begin{align}
\hat{\beta}: \pi_{1}(X,x_{1}) & \to \pi_{1}(X,x_{0}) \\
[g] & \mapsto [\alpha] * g * [\overline{\alpha}].
\end{align}$
Indeed, $\begin{align}
\hat{\beta}(\hat{\alpha}([f]))= & \hat{\beta}([\overline{\alpha}] * [f] * [\alpha]) \\
= & [\alpha] * [\overline{\alpha}] * [f] *[\alpha] * [\overline{\alpha}] \\
= & f \\
= & [\overline{\alpha}] * [\alpha] * [f] *[\overline{\alpha}] * [\alpha] \\
= & \hat{\alpha}([\alpha] * [f] *[\overline{\alpha}] ) \\
= & \hat{\alpha}(\hat{\beta}([f])).
\end{align}$
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####
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```