From what I know, a continuous function can be intuitively thought of as a mapping that does not tear but only stretches or rotates, at least in the Euclidean space. Let $X$ be the one-point union of two circles in $\mathbb{R}^2$. Does the following map give a continuous function from $X$ to itself that does not have any fixed point? I'm trying to show that $X$ does not have the fixed-point property.
Thinking of $X$ as having the subspace topology inherited from $\mathbb{R}^2$, is the operation of rotating (even if in the 'process' of rotating, the image fails to lie in $X$) still continuous?
Edit: I guess the second map isn't a continuous function from $X$ to itself. My intuition is then $X$ does have the fixed-point property because mapping the intersection point anywhere else other than to itself breaks continuity. How can I show this rigorously?
Conifold already showed you a continuous $f: X \to X$ without a fixpoint: mirror the right hand : first apply $(x,y) \to (-x,y)$ (assuming we have two circles meeting in $(0,0)$ on both sides of the $y$-axis) and then rotate the left circle over some angle. This composition has no fixpoint, clearly.
But if $h: X \to X$ is a homeomorphism it must map the intersection point to itself, because that is the unique cut-point of $X$: If $h(0) = p \neq 0$ then $h\restriction_{X\setminus\{0\}}$ is a homeomorphism of $X\setminus \{0\}$ with $X\setminus \{p\}$ and the latter is connected while $X\setminus \{0\}$ is not.
So $X$ does not have the FFP, but it does have the FFPH: the fixed point property for homeomorphisms.