Referring to G. Perelman, Proof of the soul conjecture by Cheeger and Gromoll. Given a distance-nonincreasing retraction $P$ from an open complete manifold of nonnegative curvature onto its soul $S$, one wants to prove that $P(\exp_xt\nu)=x$ for every $x\in S$ and $t\geq 0$. Suppose this is true up to $t=l$, and consider the function $$f(r)=\max\{|xP(\exp_x(l+r)\nu)|\mid x\in S, \nu\in SN_x(S)\}$$ where $SN(S)$ is the unitary normal bundle to $S$. It seems to me that the only way to interpret the notation $|xP(\exp_x(l+r)\nu)|$ is "the Riemannian distance between the two points".
But later in the proof, Perelman does the following: he has a geodesic $\gamma(u)\in S$, a parallel normal field $\nu$ along $\gamma$, and two "vertical" geodesics related to a variation around $\gamma$, namely $\sigma_{u_0}(t)=\exp_{\gamma(u_0)}t\nu$ and $\sigma_{u_1}(t)=\exp_{\gamma(u_0)}t\nu$, $t$ varying. He then proves that
the two-dimensional strip between these two $\sigma_{u_0}$ and $\sigma_{u_1}$ is flat and totally geodesic, and that all the "horizontal" geodesics $\gamma_t(u)=\exp_{\gamma(u)}t\nu, > u\in[u_0,u_1]$ are minimizing of constant length $l(\gamma|_{[u_0,u_1]})$.
Then he claims that, for small $\delta$, $$f(r-\delta)\geq |\gamma(u_1)P(\sigma_{u_1}(l+r-\delta))|\geq|P(\sigma_{u_0}(l+r)\gamma(u_1)|-|P(\sigma_{u_0}(l+r))P(\sigma_{u_1}(l+r-\delta))|\geq|P(\sigma_{u_0}(l+r))\gamma(u_1)|-|\sigma_{u_1}(l+r-\delta)\sigma_{u_0}(l+r)|\geq|P(\sigma_{u_0}(l+r)\gamma(x_1)|-|\sigma_{u_0}(l+r)\sigma_{u_1}(l+r)|-O(\delta^2).$$ Recall that $\gamma(u)=\sigma_u(0)$ and think that you can just put $l=0$ for simplicity. The first inequalities are easy applications of the triangle inequality and the fact that $P$ does not increase distances. Now consider the last inequality of the chain just written. It seems to me that he is applying the triangle inequality to the term with sign "-" and then says that one can estimate the distance $\rho(\sigma_{u_0}(l+r-\delta),\sigma_{u_0}(l+r))$ as $O(\delta^2)$. I wonder why. Shouldn't it be $O(\delta)$?
Moreover, he says that he uses in that last in equality the fact that I put under citation. Where does he use it? Does it allow to gain the $O(\delta)$? I don't think so, since in the case of $\mathbb R^2$ that distance is precisely $\delta$.
Another hypothesis was the following: he is not using the distances, but their squares. This would provide the $\delta^2$, but what about the triangle inequality? That should not be true anymore...
So I am surely missing something important. Can someone help me?
Thank you in advance.
EDIT: Question solved. One can use flatness to use Pitagoras's theorem and then that $\sqrt{1+a^2}= 1+a^2/2+O(a^2)$.