TITLE: Map from simplex to itself that preserves sub-simplices
QUESTION [10 upvotes]: I believe this may be a standard algebraic topology problem, so I apologize in advance if this belongs in stackexchange (it's not a homework problem, however, and came about in a research context).  I've got a continuous map $f$ from the $n$-simplex to itself, such that the image of every strict sub-simplex is itself.  So, each vertex gets mapped to itself, as does each edge, and so on and so forth.  Does it follow that $f$ must be surjective?
Thank you!

REPLY [2 votes]: Recently, in search for an early reference for this statement I came across this 2009 preprint by Roman Karasev: 
KKM-type theorems for products of simplices and cutting sets and measures by straight lines.
It has the statement as Lemma 1:

In the following pdf, also by Karasev: Geometry of measures: partitions and convex bodies, 2013, the statement appears as Lemma 9.2.
There is a proof, very simliar to my other answer (and to the proof in the preprint from 2009):

Karasev also provides a very nice application: Brouwer's fixed-point theorem!

I wonder whether 2009 is the first time this Lemma was written down somewhere, my feeling is that it should have been used by somebody before that time.