fundamental group of punctuated plane Let $X$ be the plane punctuated at the origin. Let $C$ be the unit circle, with each point being identified by an angle between $0$ and $2\pi$. $f$ is a function $[0,1] \rightarrow C$ so that $f(t)=2\pi t$ and $g$ is a function $[0,1] \rightarrow C$ so that $g(t)=4\pi t$. Let $h$ be the function $[0,1]\times[0,1] \rightarrow C$ so that $f(x,t)=(1-t) f(x)+t g(x)$. This function is continuous, and for every $x$, $h(x,0)=f(x)$ and $h(x,1)=1$, so $h$ conforms to the homotopy definition at < . Since $f(0)=g(0)=0$, both $f$ and $g$ are loops with the same endpoints. I know that the fundamental group of the punctuated space is $\mathbb{Z}$, so I would like to know what is flawed in my construction, which implies that all loops are homotopic and that the fundamental group is trivial.

For the fundamental group, one considers homotopies with fixed end-points, or at least homotopies of closed curves (continuous maps $h \colon S^1\times [0,\,1] \to X$ leave less room for mistakes).

That means in your homotopy you'd need $h(0,t) = h(1,t)$ for all $t \in [0,\,1]$.

Since $f$ winds around the origin only once, and $g$ twice, that condition cannot be satisfied, and for your $h$, you have $h(0,t) = h(1,t)$ only for $t = 0$ and $t = 1$.