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You can read in Wikipedia the following:

Given a space X and a loop $\alpha\colon S^1 \to X$ representing an element of the fundamental group of $X$, we can form the mapping cone $C_α$. The effect of this is to make the loop $α$ contractible in $C_α$, and therefore the equivalence class of $α$ in the fundamental group of $C_α$ will be simply the identity element.

So we have $[\alpha (x)]=[(x,1)]$. Does it mean, that $\alpha$ is homotopic to the constant path in $[(x,0)]$ by just moving the loop continuously to the peak of the cone? Or why is the equivalence class of $\alpha$ the identity element?

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  • $\begingroup$ If we look at the image of $\alpha$ in $C_\alpha$, then we can just slide $\alpha$ over the top of the cone like you said, so you're right, $\alpha$ would be null-homotopic so it's equivalence class would be the identity element in $\pi_1(X)$. This idea is generalized a bit in the construction of a Postnikov tower for $X$ in higher homotopy theory which you can now look forward to! $\endgroup$ – Jack Davies Apr 23 '15 at 8:14
  • $\begingroup$ Yes : the mapping cone $C_\alpha$ is obtained by pasting a copy of $C(S^1) \cong D^2$ to $X$ by the identifications via $\alpha$ at the base of the cone. Then the loop $\alpha$ can be path-homotoped across the disk to the constant map at the basepoint. This means that $[\alpha]$ is just the identity element in $\pi_1(X)$. $\endgroup$ – Balarka Sen Apr 23 '15 at 8:14

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