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Let $X$ be a space and $L$ be a subspace of $X$ that is contractible. Then is $X/L$ homotopy-equivalent to $X$ itself? it doesn't seem trivial to me at al...If so, how to show it rigorously? Could anyone help me?

Hamou
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John Ralph
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    This is shown at the very start of Hatcher. Chapter 0, page 11. – Harry Wilson Oct 11 '14 at 10:15
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    You would probably want $(L,X)$ to satisfy the homotopy extension property. – user101036 Oct 11 '14 at 10:36
  • You would certainly want $(L,X)$ to satisfy some conditions, of which the HEP is the usual one. The result is a special case of the gluing theorem for homotopy equivalences: see this question http://mathoverflow.net/questions/96071/general-gluing-theorem-for-adjunction-spaces – Ronnie Brown Oct 11 '14 at 14:58
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    As an example of where this is not true. Take $X=S^1$ and for some point $x\in X$, let $L=X\setminus{x}$. The space $X/L$ here is the Sierpinski space which is contractible, so not homotopy equivalent to $S^1$. – Dan Rust Oct 11 '14 at 17:35

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