Let $X$ and $Y$ be topological spaces. Assume $Y$ is [contractible](http://en.wikipedia.org/wiki/Contractible_space) (hence, path- connected). 


Let $f,g: X \to Y$ be continuous maps. At any fixed $x\in X$, there is a path $P\_x: [0,1]\to Y$ from $f(x)$ to $g(x)\in Y$ such that $P\_x(0)=f(x)$ and $P\_x(1)=g(x)$


We define $F: X\times [0,1]\to Y$ by $F(x,t) = P\_x(t)$. Now, $$F(x,0)= P\_x(0)=f(x), F(x,1)= P\_x(1)=g(x)$$ for any $x\in X$ (This is why $F$ is homotopy-like). Clearly, $F$ is continuous at time $0$ and at time $1$. Is $F$ necessarily continuous at any time $t$?


Thanks in advance for any help.



Continuity of a homotopy-like function
This proof cannot work since then every two maps $X \to Y$, where $Y$ is path-connected, are homotopic - which is false. On the other hand, if $Y$ is contractible, then every map $X \to Y$ is homotopic to a constant map (since this true for the identity $X \to X$), thus every two maps $X \to Y$ are homotopic.


Even if $Y$ is contractible, in your proof the paths $P\_x$ may be choosen so arbitrarily that $F$ is not continuous. Indeed, there are maps $F : [0,1] \times [0,1] \to \mathbb{R}$ such that $F(-,0)$, $F(-,1)$ and all $F(t,-)$ are continuous, but $F$ is not.


