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A

As other have said, in the one dimensional case at least, you can suppose that the volatility is constant. Then the solution of the SDE is nothing else that than a solution of the integral equation $$X(t) = \int_0^t \mu(X_s) ds + \sigma W_t \qquad \forall t \in [0,T].$$ You can then check that if $\mu(\cdot)$ is a Lipschitz function, say, then the function $\Psi$ that maps $(W_t)_{t \in [0,T]}$ to the solution of the above integral equation is continuous (Gronwall Lemma) on $C([0,T],\mathbb{R})$ with the supremum norm. Hence you can indeed write $X = \Psi(W)$ and see the path $(X_t)_{t \in [0,T]}$ as a 'deformation' of the Brownian path $(W_t)_{t \in [0,T]}$. The function $\Psi: C([0,T],\mathbb{R}) \to C([0,T],\mathbb{R})$ is sometimes called the 'Ito map' in the literature.
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A other have said, in the one dimensional case at least, you can suppose that the volatility is constant. Then the solution of the SDE is nothing else that a solution of the integral equation $$X(t) = \int_0^t \mu(X_s) ds + \sigma W_t \qquad \forall t \in [0,T].$$ You can then check that if $\mu(\cdot)$ is a Lipschitz function, say, then the function $\Psi$ that maps $(W_t)_{t \in [0,T]}$ to the solution of the above integral equation is continuous (Gronwall Lemma) on $C([0,T],\mathbb{R})$ with the supremum norm. Hence you can indeed write $X = \Psi(W)$ and see the $(X_t)_{t \in [0,T]}$ as a 'deformation' of the Brownian path $(W_t)_{t \in [0,T]}$. The function $\Psi: C([0,T],\mathbb{R}) \to C([0,T],\mathbb{R})$ is sometimes called the 'Ito map' in the literature.