= Solution
Let $P$ be the homotopy fiber of the map representing
$$
\operatorname{Sq}^1:K(\mathbb F_2,1)\longrightarrow K(\mathbb F_2,2).
$$
The long exact sequence of <homotopy groups> shows that every $\pi_i(P)$ except $\pi_1(P)$ vanishes and that
$$
0\longrightarrow\mathbb F_2\longrightarrow\pi_1(P)
\longrightarrow\mathbb F_2\longrightarrow0.
$$
Hence $P$ is either $K(\mathbb F_2\oplus\mathbb F_2,1)$ or $K(\mathbb Z/4,1)$.
The extension is classified by the degree-two class represented by the original map. If $t\in H^1(K(\mathbb F_2,1);\mathbb F_2)$ is the standard generator, that class is
$$
\operatorname{Sq}^1t=t^2\ne0
$$
in $H^*(\mathbb{RP}^{\infty};\mathbb F_2)=\mathbb F_2[t]$. It therefore classifies the non-split extension, whose middle group is $\mathbb Z/4$. Thus
$$
P\simeq K(\mathbb Z/4,1).
$$
Solved by gpt-5.6-sol high.
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