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Let $X$ be a simplicial set. Let $X\to \Delta^n$ be a right fibration (has the right lifting property with respect to right horn inclusions), and let $$\Delta^{\{n-i\}}\hookrightarrow \Delta^{\{n-i,\dots, n\}}\hookrightarrow \Delta^n$$ (for a fixed $i: 0\leq i\leq n$) be the obvious inclusion maps.

Then why is the induced map:

$$X\times_{\Delta^n} \Delta^{\{n-i\}} \hookrightarrow X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}}$$

a deformation retract? It's not like we can apply Whitehead's theorem, since $\Delta^n$ is not a Kan complex in general.

(This is from the end of the proof of proposition of HTT. The statement should be true out of the context in the book with the hypotheses I've given, but if not, there's the source.)

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In topology, if a map $Y\to B$ is some sort of fibration then you would think that $B$ being equivalent to a subspace $A$ would imply that $Y$ is equivalent to $Y_A=Y\times_BA$. If fibration means Serre fibration and equivalent means weakly, then you might want to use homotopy groups and the Whitehead Theorem. But what if we try to prove directly that a deformation retraction of $B$ to $A$ would yield a deformation retraction of $Y$ to $Y_A$? Then we will want to assume that $Y\to B$ has the 'relative homotopy lifting property' for the pair $(Y,Y_A)$, that is, the right lifting property with respect to the inclusion $Y\times 0\cup Y_A\times I\to Y\times I$.

I suppose that this it how it is in what you are looking at: Any right fibration of simplicial sets has the RLP w.r.t. $K\times \lbrace 1\rbrace \cup L\times \Delta^1\to K\times \Delta^1$ for every $L\subset K$ (but not w.r.t. $K\times \lbrace 0\rbrace \cup L\times \Delta^1\to K\times \Delta^1$), and this is good for lifting the sort of deformation retraction that exists from a simplex to its zeroth vertex.

Something like that.

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Harry, I'm not sure how you are using the word 'contractible' for simplicial sets. And, come to think of it, I'm not sure how I would use it. But Lurie is definitely asserting that the inclusion of that vertex into that i-simplex 'is a deformation retract', because you can take X to be the n-simplex. Note that, even though $\Delta^n$ is not fibrant, there is a map $\Delta^n\times \Delta^1\to\Delta^n$ that can be called a homotopy relating the identity to a constant map to a vertex. First or last vertex dictates which end of the homotopy the identity is at. – Tom Goodwillie Jun 30 '10 at 10:33
@Tom, I think you need to compose two such homotopies if your vertex is neither the first nor the last (a self-map of $\Delta^n$ is a monotone map on the set of vertices, and there is a unique homotopy from one monotone map f to another g as you describe iff $f(x) \leq g(x)$ for all x). – Tyler Lawson Jun 30 '10 at 14:36
@Tyler: Yes. I did not make that explicit because I was coming up against the length limit for a comment. – Tom Goodwillie Jun 30 '10 at 14:58
Ah, my apologies. – Tyler Lawson Jun 30 '10 at 16:02
Such an orgy of politeness! No apology necessary. – Tom Goodwillie Jun 30 '10 at 20:43

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