* notation
@ 2017-05-04 18:23 Thierry Coquand
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From: Thierry Coquand @ 2017-05-04 18:23 UTC (permalink / raw)
To: homotopy Type Theory
I was told that the notation in my previous message may be confusing: I write
m_c : X -> D_c(X)
with X implicit argument.
With X explicit argument, the notation would be
m_c X : X -> D_c(X)
and condition of being idempotent is that D_c(m_c X) and m_c (D_c X) are
path equal.
Since D_c(X) = X^{F(c)} this can also be expressed as the fact that
X^{pi_1} and X^{pi_2} : X^{F(c)} -> X^{F(c) x F(c)}
are path equal (note that F(c) is not required to be fibrant).
Another way to state it as a proposition is that the multiplication of the monad
X^{diagonal} : X^{F(c) x F(c)} -> X^{F(c)}
is an equivalence.
Thierry
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