Homotopic maps which aren't relative homotopic 
Can anyone give an example of two continuous maps, let say $f,g \colon X \to Y$, such that the set $A =\{ x \in X \mid f(x) = g(x) \}$ is not empty, the two maps are homotopic but there's no homotopy between these function relative to any non empty subspace of $A$?

 A: Let $X = S^1$, let $p \in S^1$, and let $Y = S^1 \times [0;1]$ with $(p,0)$ and $(p,1)$ identified.
Then let $f(x) = (x,0)$ and $g(x) = (x,1)$. Clearly, $h(x,t) = (x,t)$ is an homotopy from $f$ to $g$, but there is no homotopy relative to $p$.
Suppose $h : S^1 \times [0;1] \to Y$ is an homotopy from $f$ to $g$ such that $h(p,t) = (p,0)$ forall $t$.
Let $q$ be a point of $S^1$ distinct from $p$, and $A$ be the arc-connected component of $(q,0)$ in $(S^1 \times [0;1]) \setminus h^{-1}(\{(p,0)\})$ (this set contains $(q,0)$ because $h(q,0) = f(q) = (q,0) \neq (p,0)$) and define $\tilde h : S^1 \times [0;1] \to Y$ by $\tilde h(x,t) = h(x,t)$ for $(x,t) \in A$, and $\tilde h(x,t) = (p,0)$ for $(x,t) \notin A$. 
$\tilde h$ is still continuous : The arc-connected components of an open set are all open, so if $h(x,t) \neq (p,0)$, there is an open neighbourhood of $(x,t)$ in its arc-connected component, and in both cases (wether it is the one of $(q,0)$ or not), $\tilde h$ is continuous there.
If $h(x,t) = (p,0)$, since $h$ is continuous, for every neighbourhood $U$ of $(p,0)$ there is a neighbourhood $V$ of $(p,0)$ such that $h(V) \subset U$. Since $\tilde h(V) \subset h(V) \cup \{(p,0)\} = h(V) \subset U$, that same neighbourhood works to show $\tilde h$ is continuous at $(x,t)$.
If $(q,1) \notin A$, then $(x,1) \notin A$ forall $x \in S^1$, so $\tilde h$ is an homotopy between $f$ and the constant map $(p,0)$, which is impossible because $f$ does one loop around the cylinder and the constant map doesn't.  
Thus $(q,1) \in A$. We have a path $\gamma : [0;1] \to S^1 \times [0;1] = (\gamma_1,\gamma_2)$ such that $\gamma(0) = (q,0), \gamma(1) = (q,1)$, and $h \circ \gamma$ doesn't touch $(p,0)$. In particular, $\gamma$ doesn't touch any $(p,t)$, so its first coordinate, $\gamma_1$, has values in $S^1 \setminus \{p\}$.
If we further identify $S^1 \times \{0\}$ with $S^1 \times \{1\}$ (in $Y$) to get a torus, $h \circ \gamma$ is a loop from $(q,0)$ to $(q,1)$, which loops exactly once around the second coordinate. But in fact, it is also a loop homotopic to the null loop :  
Pick a continuous map $\kappa : (S^1 \setminus \{p\}) \times [0;1] \to S^1$ such that $\kappa(x,0) = x$ and $\kappa(x,1) = p$ forall $x$, and define $\theta : [0,1] \times [0,1]$ by $\theta(t,s) = h(\kappa(\gamma_1(t),s),\gamma_2(t))$. It is a loop homotopy between $h \circ \gamma$ and $p$ :
$\theta(0,s) = h(\kappa(q,s),0) = f(\kappa(q,s)) = (\kappa(q,s),0)$, which is identified to
$\theta(1,s) = h(\kappa(q,s),1) = g(\kappa(q,s)) = (\kappa(q,s),1)$.
$\theta(t,0) = h(\kappa(\gamma_1(t),0),\gamma_2(t)) = h(\gamma_1(t),\gamma_2(t)) = h \circ \gamma(t)$
$\theta(t,1) = h(\kappa(\gamma_1(t),1),\gamma_2(t)) = h(p,\gamma_2(t)) = (p,0)$  
This is impossible too, so there can be no such $h$.
A: Actually I recently come back to this problem and I've found myself another counterexample. 
Consider the space 
$$X= \{ (x,y) \in \mathbb R^2| (x \ne 0 \land y/x \in \mathbb Q) \lor x=0\}$$
this is the space formed by the straight lines with rational slope.
This space can be deformation retracted on the point $(0,0)$, I'm going to call $p \colon X \to X$ the constant map sending $X$ in this point, and so the identity is relative homotopic to $p$.
The point $(0,0)$ is the only point to which the whole space can deformation retract.
Clearly every other constant map is homotopic to this map, so every other constant map is homotopic to the identity. 
Any constant map have a unique fixed point, which clearly is fixed by the identity too. The identity cannot be homotopy equivalent to any constant map 
relatively to the fixed point, with the exception of $p$, because otherwise the relative homotopy would be a deformation retraction of $X$ on this other point, which is not possible for said above.
