# Where is the loophole in this homotopy argument?

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 ?

• 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. – jgon May 26 '20 at 7:08
• Ohh..I get it now. The statement is not true for path homotopy in general. Thanks @jgon – Ganesh Gani May 26 '20 at 7:20
• Yep, exactly. Np :) – jgon May 26 '20 at 7:21
• "Where is the loophole..." I see what you did there – SolveIt May 26 '20 at 7:42