For any integer $n\ge 0$, let $[n]$ be the totally ordered set of the integers $k$ such that $0\leq k\leq n$, with the usual order relation.
Here are the last lines from the proof of Proposition 1.1.3.11 of Kerodon (with $n,n_-,n_+$ nonnegative integers):
the map of partially ordered sets $[n]\to [n_-]\times [n_+]$ is a monomorphism, so that $n\leq n_-+n_+$.
I can't figure out how to justify this claim: it doesn't seem to rely on a specific result, it looks more like a basic fact; however the only thing I can think of is that the product the monomorphism hypothesis guarantees that the cardinality of $[n]$ (i.e. $n+1$) is less than or equal to that of $[n_-]\times [n_+]$. But isn't the latter cardinality $n_- n_+$, instead of $n_- +n_+$? Honestly the indented claim doesn't even seem true to me, but I don't understand how the rest of the context would influence the interpretation of this claim. Would you help me to see its sense? Thank you for your patience
The poset $[n_-]\times [n_+]$ looks like a grid with $n_-$ rows and $n_+$ columns, with $(a,b)\leq (a',b')$ iff $a\leq a'$ and $b\leq b'$, i.e., $x\leq y$ iff the row of $y$ is above or equal to the row of $x$ and the column of $y$ is to the right or equal to the column of $x$.
We have an injective order-preserving function $f$ from $[n]$ into this grid. So for all $0\leq k<n$, $f(k) < f(k+1)$, and this implies that either (a) $f(k+1)$ is at least one row above $f(k)$ or (b) $f(k+1)$ is at least one column to the right of $f(k)$. Now in the $n$ steps from $0$ up to $n$, case (a) can only happen at most $n_-$ many times, and case (b) can only happen at most $n_+$ many times. So $n\leq n_-+n_+$.