I$\DeclareMathOperator\Top{Top}$I would like to compute $\pi_3Top(4)$$\pi_3\Top(4)$ and $\pi_4Top(4)$$\pi_4\Top(4)$. It is known that $Top(4)/O(4) \rightarrow Top/O$$\Top(4)/O(4) \rightarrow \Top/O$ is 5-connected and $$ \pi_k(Top/O) = \begin{cases} 0, & k = 0,1,2,4; \\ \mathbb{Z}/2, & k = 3 . \end{cases} $$$$ \pi_k(\Top/O) = \begin{cases} 0, & k = 0,1,2,4; \\ \mathbb{Z}/2, & k = 3 . \end{cases} $$ The long exact sequence of homotopy groups for fibre sequence $Top(4)/O(4) \rightarrow BO(4) \rightarrow BTop(4)$$\Top(4)/O(4) \rightarrow BO(4) \rightarrow B\Top(4)$ is: $$\pi_5Top(4)/O(4) \rightarrow \pi_4 O(4) \rightarrow \pi_4 Top(4)\rightarrow \pi_4Top(4)/O(4) \rightarrow \pi_3 O(4) \rightarrow \pi_3 Top(4) \rightarrow\pi_3Top(4)/O(4) \rightarrow \pi_2 O(4)$$$$\pi_5\Top(4)/O(4) \rightarrow \pi_4 O(4) \rightarrow \pi_4 \Top(4)\rightarrow \pi_4\Top(4)/O(4) \rightarrow \pi_3 O(4) \rightarrow \pi_3 \Top(4) \rightarrow\pi_3\Top(4)/O(4) \rightarrow \pi_2 O(4)$$ By plugging in the known groups ($SO(4) \cong S^3 \times \mathbb{R}P^3$) we get two exact sequences: $$\pi_5Top(4)/O(4) \rightarrow \mathbb{Z}/2 \times \mathbb{Z}/2 \rightarrow \pi_4 Top(4) \rightarrow 0$$$$\pi_5\Top(4)/O(4) \rightarrow \mathbb{Z}/2 \times \mathbb{Z}/2 \rightarrow \pi_4 \Top(4) \rightarrow 0$$ $$0 \rightarrow \mathbb{Z} \times \mathbb{Z} \rightarrow \pi_3 Top(4) \rightarrow \mathbb{Z}/2 \rightarrow 0$$$$0 \rightarrow \mathbb{Z} \times \mathbb{Z} \rightarrow \pi_3 \Top(4) \rightarrow \mathbb{Z}/2 \rightarrow 0$$ I can also show that $\pi_4Top(4)$$\pi_4\Top(4)$ is at least $\mathbb{Z}/2$ by comparing to the same long exact sequence for $O(5), Top(5)$$O(5), \Top(5)$, but it still may be $\mathbb{Z}/2 \times \mathbb{Z}/2$. Is there a way to finish the computation?