Show that if $X$ is path connected then every path $f:I\to X$ is homotopic to a constant path $g(t)=x$ Show that if $X$ is path connected then every path $f:I\to X$ is homotopic to a constant path $g(t)=x$. Does this mean show that there is an $x$ (or even for all $x$) such that this is true for all paths $f$, or to show for each $f$, there is an $x$ such that this is true? I know that two paths are homotopic if there exists a continuous map $H(s,t):I\times I\to X$ with $H(s,0)=f(s)$ and $H(s,1)=g(s)$. I think I'm just misunderstanding the question. If we take an annulus, which is path connected, and consider a loop around the inner circle, then how is this loop, which is a path, homotopic to any constant loop? Could somebody explain what the question is asking since I can't make any progress with it until I understand. I thought googling this question would tell me something and nothing came up. Is the claim false or am I misunderstanding something?
 A: This can work (to assuage your doubts) as the homotopy is a "free" one, where the homotopy does not have to preserve any fixed point, as in the fundamental group, which is why the torus or circle don't work as counterexamples..
If $f$ is a path, we can define $H: [0,1] \times [0,1] \to X$ by $H(s,t) = f(s(1-t))$ which is clearly continuous, and for $t=0$ we just have the path $f$ and for $t=1$ a constant map with constant value $f(0)$. We could have an arbitary $x$ for the constant because in path connected space $X$ any two constant maps are homotopic (and homotopy is an equivalence relation).
A: Well, I think that the quastion could be wrong, because "path connected" is not enough to ensure that any path is homotopic to a point. As a counterexample, a loop in the circle $\mathbb{S}^1$ (with its usual topology) is not homotopic to a point.
The question that arises then is: What is the definition of "path connected"? Linking it with the one in your question, I would write:
Definition:
Let $(X,\tau_X)$ be a topological space. Then $X$ is path connected if $\forall x,y\in X$ two points in $X$, $\exists f\in Mor_{Top}(I,(X,\tau_X))$ such that $f(0)=x$ and $f(1)=y$.
Conversely, both $x$ and $y$ can be understood as constant paths. In that case, $f$ won't be a path, but a homotopy.
Let $x\in Mor_{Top}(I,(X,\tau_X))$, $y\in Mor_{Top}(I,(X,\tau_X))$, such that $x(t)=x, y(t)=y\; \forall t\in I$. If $(X,\tau_X)$ is path connected, there must exist $H\in Mor_{Top}(I\times I,(X,\tau_X))$ such that $H(0,t)=x(t)=x$ and $H(1,t)=y(t)=y$. Essentially, $H(s,t)$ is the same thing as $f(t)$, but defined as a different object.
Note that the topologies are necessary to check if the given morphisms are continuous, as they have to be, because they are morphisms of Top category.
