I was reading Hall's book on Lie groups. After defining Connected Lie groups he stated and proved a proposition :
If $G$ is a matrix Lie group then the component of $G$ containing identity is a subgroup of $G$.
For proving the proposition first he proves that if $A$ and $B$ belongs to the component (which contains identity $I$) we can take the products of the continuous paths connecting $I$ to $A$ and $I$ to $B$ in such a way that the new path will be from$I$ to $AB$.Thus there exists a continuous path from $I$ to $AB$.So $AB$ lies in that component.
Next he has to prove that if $A \in $ component then $A^{-1}$ also lies in that component.For that he argues like this : Since $A$ belongs to this component we can find a continuous path from $I$ to $A$.Let it be $A(t)$.Then $A(t)^{-1}$ is a continuous path from $A^{-1}$ to $I$.This argument I could not understand because
according to my knowledge continuous path $A(t)$ means a continuous function $A(t)$ from $[0,1]$ to $G$ (in this case $A(0)=I $ and $A(1)=A$).
Then $A(t)^{-1}$ will be a map from $G$ to $[0,1]$ such that $A^{-1}(I)=0$ and $A^{-1}(A)=1$. It will not connect $I$ and $A^{-1}$.So how to handle this problem ?
I think you're just confusing the notation of inverse for a matrix versus inverse of a mapping. Recall that in a Lie group $G$ and given $A \in G$ the map $A \to A^{-1}$ is differentiable. To clarify the notation, let $f:[0,1] \to G$ be differentiable such that $f(0) = I$ and $f(1) = A$. Then take the inverse $A^{-1}$ and define $g(t) = A^{-1}f(t)$. This gives a differentiable map starting at $A^{-1}$ and ending at $I$.