Simple Connectedness of [0,1]^k

[quasar987]

I am wondering if k-cubes preserve simple connectedness. I.e., is the image of [0,1]^k by a continuous function c:[0,1]^k-->R^n simply connected?

Wiki has shown me that continuous maps do not take simple connected sets to simple connected sets. For instance exp:C-->C takes C to C\{0}.

But is it true for [0,1]^k ?

 

[morphism]

What about c(x) = (cos(2pix), sin(2pix)) for x in [0,1]?

 

[quasar987]

Yeah. :P

 

[quasar987]

On the other hand, if M is a k-manifold imbedded in R^n and c:[0,1]^k-->M is a k-cube homeomorphic onto its image, then c([0,1]^k) is simply connected because it has a trivial fundamental group. Correct?