I recently had a problem:
Show that if $X$ is contractible, and $Y$ is path-connected, show that the homotopy class $[X,Y]$ has a single element.
I have been able to prove this (I think) in a fairly "standard way":
Take $f,g:X \to Y$ and homotopy $F(x,t)$ between $i_x$ and constant function at $x_0$. Then consider the path $\alpha$ from $f(x_0)$ to $g(x_0)$.
Then just define $H:X \times I \to Y$ by: $$H(x)= \begin{cases} f(F(x,2t)) & t \in [0,1/2] \\ \alpha(2t-1) & t \in [1/2,1] \end{cases}$$
I can visually see this a bit. My question then is why we even need to define the first part of the homotopy in terms of a constant function? So, what is wrong with the following "proof":
So if we had $x \in X$. Then $f(x)=y_1$ and $g(x)=y_2$. But then for each $x \in X$, there is a unique path $\beta_{x}:I \to Y$ from $y_1$ to $y_1$. Then just define homotopy $H: X \times I \to Y$ by $H(x,t)=\beta_{x}(t)$.
Is the problem that each path cannot be guaranteed to have the same $t$ value? Or instead that each particular path does not apply to all $x \in X$?
In other words: does the desired result necessarily depend on the contractibility of $X$?
This is an aside: is it true that in general, a contractible space has the trivial fundamental group?
In order to avoid a long discussion, let me clarify this: the first proof is incomplete. Let $\alpha$ is a path from $f(x_0)$ to $g(x_0)$, then your $H$ is a homotopy from $f$ to a constant map $x\mapsto g(x_0)$. On the other hand, $g\circ F$ is another homotopy from $g$ to the constant map $x\mapsto g(x_0)$. Then you use the fact that being homotopic is an equivalence relation.
For the second "trial", you usually cannot pick an arbitrary path $\beta_x(t)$ from $f(x)$ to $g(x)$ for all $x$ and hope that the resulting homotopy $(x,t)\mapsto \beta_x(t)$ will be continuous. Note that you never used the contractibility assumption.