-3
$\begingroup$

Let $C$ be a model category. Recall that a morphisme $f : X \to Y$ is called a homotopy monomorphism if the diagonal $X \to X \times^h_Y X$ induces an isomorphism in the homotopy category. Suppose that one has morphisms $i : X \to Y$, $p : Y \to X$ such that $p i = 1_X$. Is $i$ then a homotopy monomorphism (and dually $p$ a homotopy epimorphism) ?

$\endgroup$
2
  • $\begingroup$ Why the downvote ? $\endgroup$ Mar 24, 2015 at 19:32
  • $\begingroup$ Because you didn't show any effort...(I'm not the one who downvoted, but I'm sure whoever did did so for this reason most likely) $\endgroup$ Mar 24, 2015 at 20:25

1 Answer 1

1
$\begingroup$

This is false even in $\mathbf{sSet}$. For instance, take any non-contractible connected simplicial set $X$ and any morphism $i : \Delta^0 \to X$. There is a unique morphism $p : X \to \Delta^0$ and, of course, $p \circ i = \mathrm{id}$. But $i : \Delta^0 \to X$ cannot be a homotopy monomorphism: this happens if and only if the loop space of $X$ is (weakly) contractible, and since $X$ is connected, that happens if and only if $X$ is (weakly) contractible.

$\endgroup$
2
  • $\begingroup$ Thanks ! Might you suggest a correct definition of "homotopy split monomorphism" ? Does such a notion make sense ? $\endgroup$ Mar 24, 2015 at 20:26
  • $\begingroup$ The trouble, really, is in "monomorphism", not "split". Monomorphisms and epimorphisms in $(\infty, 1)$-categories are quite subtle things. $\endgroup$
    – Zhen Lin
    Mar 24, 2015 at 20:47

Not the answer you're looking for? Browse other questions tagged .