Suppose $f,g:[0,1]\to [0,1]$ are two maps then we have that $f$ is homotopic to $g$ since $[0,1]$ is a convex space and if $h:[0,1] \to X$ is a path in an arbitrary space $X$, it implies $h \circ f$ homotopic to $h \circ g$ since homotopy is invariant under composition.

Now, if I take $f=Id_{[0,1]}$ and $g=0$, then this implies $h \circ f = h$ is homotopic to $h \circ g = c_{h(0)}$ i.e any path in $X$ is homotopic to a constant map which is not true in general.

What am I missing here ?

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    $\begingroup$ It is true that any path in $X$ is homotopic to the constant map as maps from the interval to $X$, but you're probably thinking of path homotopies, which require the endpoints to remain fixed throughout the homotopy, and therefore restrict what homotopies are allowed. $\endgroup$ – jgon May 26 '20 at 7:08
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    $\begingroup$ Ohh..I get it now. The statement is not true for path homotopy in general. Thanks @jgon $\endgroup$ – Ganesh Gani May 26 '20 at 7:20
  • $\begingroup$ Yep, exactly. Np :) $\endgroup$ – jgon May 26 '20 at 7:21
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    $\begingroup$ "Where is the loophole..." I see what you did there $\endgroup$ – SolveIt May 26 '20 at 7:42

To close the question: any path in a space is contractible, which implies that all paths into a path-connected space are homotopic. However, this is not true if we require the endpoints to be fixed during the homotopy.


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