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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 $A$?

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Did you mean "relative to every non empty subspace of $A$"? –  user31373 Sep 24 '12 at 1:08
@LVK No I mean $X$, but if you want you can consider $A$ to be the set $\{x \in X \mid f(x) = g(x)\}$. –  Giorgio Mossa Sep 24 '12 at 6:55
I don't understand the question as it stands. Could you please clarify what you want? I gather $f$ and $g$ should be freely homotopic, agree on some non-empty subset, but I don't understand what follows. –  Olivier Bégassat Sep 30 '12 at 5:06
@OlivierBégassat I've slightly changed the question, I hope now the question is more clear. –  Giorgio Mossa Sep 30 '12 at 10:36
If $f,g$ are homotopic, they are also homotopic relative to $\emptyset$. I assume you want $f,g$ homotopic, $A$ not empty and for all $p\in A$, $f,g$ are not homotopic relative to $\{p\}$? –  Hagen von Eitzen Sep 30 '12 at 16:17
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2 Answers

up vote 4 down vote accepted

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 $h$.

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Sadly in this case the set $\{x \in X| f(x) = g(x)\}$ is empty, so this doesn't answer the question. –  Giorgio Mossa Sep 30 '12 at 17:27
$(p,0)$ and $(p,1)$ are identified in $Y$ so $A=\{p\}$. –  J.K.T. Sep 30 '12 at 17:43
My bad, so now the two maps are maps from $S^1$ to the torus and are the same map, am I correct? (in this case, I suppose that the two maps are homotopic relatively to the whole space........) –  Giorgio Mossa Sep 30 '12 at 17:48
No, $p$ is only one point of the circle, $Y$ is not the torus, it's a cylinder where the end circles are glued at one point. I think I could have picked a torus minus one point for $Y$, maybe it would have been clearer –  mercio Sep 30 '12 at 18:01
Ah, ok thanks now I get it, could you add the details of the non existence of the relative homotopy, so I can accept the answer? –  Giorgio Mossa Sep 30 '12 at 18:26
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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.

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