Let $K$ be a number field. $a \in K$ is said to be totally positive if $a^{\sigma}$ is positive for all real embeddings $\sigma$ of $K$. A principal ideal of $\mathcal{O}_K$ is said to be totally positive if it can be written in the form $a\mathcal{O}_K$ for totally positive element $a$ in $K$. The totally positive principal ideals form a subgroup $P_{K}^{+}$ of $P_{K}$, the principal fractional $\mathcal{O}_K$-ideals. We define the narrow ideal class group of $K$ to be $\text{Cl}^{+}_K=I_K/P^{+}_K$, where $I_K$ is the group of fractional ideals in $\mathcal{O}_K$.
Show that $\#\text{Cl}^{+}_K \leq 2^r \# \text{Cl}_K$, where $r$ is the number of real embeddings of $K$.
I know that ideal class group is a quotient of narrow ideal class group by isomorphism theorem $$\text{Cl}_K=I_K/P_K \simeq \text{Cl}_K^{+}/(P_K/P^+_K),$$ so the narrow class number is the multiple of the class number. But I don't know why the ratio should be a power of $2$. Considering there are $r$ real embeddings, I suppose the chance of having $\sigma(a)>0$ for all real embeddings is kinda related to $2^r$, but it's not clear to me how to appropriately states it.
The best formulation to state and prove this is the exact sequence: $$1\longrightarrow U/U^+ \longrightarrow K^\times / K^+ \longrightarrow C^+ \longrightarrow C \to 1$$ where $C^+$ is the narrow class group, $C$ is the class group, $U^+$ are totally positive units, $K^+$ are totally positive elements of $K$. This is not difficult to see by writing down maps explicitly.
Therefore $$\frac{[U:U^+] |C^+|}{[K:K^+] |C|} = 1 \iff |C^+|=h\frac{2^r}{[U:U^+]}$$ Finally note that for every $u\in U$, $u^2\in U^+$, so denominator is a power of $2$.