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I am just starting to read about algebraic topology, and I wonder whether homotopy depends on function or the image. According to Munkres' definition, two continuous function $f,g:[0,1]\to Y$ are said to be homotopic if there exists a continuous map $F:[0,1]\times[0,1]\to Y$ such that

$$F(0,x)=x_0$$ $$F(1,x)=x_1$$ $$F(x,0)=f(x)$$ $$F(x,1)=g(x)$$

Now if $f:[0,1]\to Y$ is a given continuous function, and $g:[0,1]\to f(X)$ is a surjective continuous function with the same end points as $f$. Is it necessary for $f$ and $g$ homotopic?

Y.H. Chan
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  • What exactly are you asking? "If $g: X \rightarrow Y$ is continuous, and the image of $g$ is equal to the image of $f$, then $g$ and $f$ are homotopic." (?) – D_S Nov 23 '15 at 04:03
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    The homotopy between $f,g$ as you defined it is ill-defined. For two maps to be homotopic the the maps should have the same domain and codomain. – Hamed Nov 23 '15 at 04:05
  • Why? Continuity is independent of the choice of codomain. – D_S Nov 23 '15 at 04:07
  • @Hamed: Sorry for that. I just rewrite my definition and so that $f$ and $g$ agree on the same end point. What I want to ask is that whether $f$ and $g$ are homotopic if the are representing the same curve. – Y.H. Chan Nov 23 '15 at 04:07
  • @Hamed I assume he wants the codomain of both functions to be $Y$, but he is emphasizing that $f(X)=g(X)$ with the notation. – Mario Carneiro Nov 23 '15 at 07:18
  • http://www.maths.kisogo.com/index.php?title=Continuity_and_non-surjective_functions continuity on non-surjective functions isn't bad (first draft page with proof though) – Alec Teal Nov 23 '15 at 07:54

2 Answers2

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No. Consider the case of the unit circle $S_1=\mathbb R/\mathbb Z$. The map $f(x)=x$ winds around the circle once, where $g(x)=x+x$ winds around the circle twice. They have the same image and the same endpoints, but are not homotopic.

A somewhat more general example is to take any curve not homotopic to the trivial curve. Then, construct a new curve which traverses that curve once, and then traverses it in reverse. The former isn't homotopic to the trivial curve, but the latter is.

Milo Brandt
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    What's worse, if the spaces involved are nice enough (say, finite CW complexes) every map is actually homotopic to a surjective map. –  Nov 23 '15 at 04:08
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No. Consider the paths $$ f: [0, 1] \to S^1 : t \mapsto (\cos 2 \pi t, \sin 2 \pi t)\\ g: [0, 1] \to S^1 : t \mapsto (\cos 4 \pi t, \sin 4 \pi t). $$ These are both surjective onto the unit circle $S^1$, and have the same starting and ending points, but are not homotopic as paths in $S^1$. (They are homotopic as maps into $\mathbb R^2$, though!)

John Hughes
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