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I came across this assertion:

There is an epimorphism $X \overset{f}\to Y\;$ in Top such that the homotopy class $X \overset{\tilde{f}}\to Y\;$ of $f$ is not an epimorphism in hTop.

Then, by way of elaboration, the text briefly refers to "the covering projection of the real line onto the circle, defined by: $x \mapsto e^{ix}$."

I interpret this to mean that $f\;$ is $x \mapsto e^{ix}$, so $X = \mathbb{R}$, and $Y = S^1$. If this interpretation is correct, I can see that this $f:\mathbb{R}\to S^1$ is an epimorphism in Top, but I still don't see how to complete the alluded-to counterexample.

More specifically, if I understand the situation correctly, the claim above amounts to asserting that there exist a topological space $(Z, \tau\,)$ and a pair of Top morphisms $g_1, g_2:S^1 \to Z$ such that all the following hold:

  1. $\tilde{g_1}\;\;\tilde{\scriptstyle\circ}\;\;\tilde{f}\,\;=\;\;\tilde{g_2} \;\;\tilde{\scriptstyle\circ}\;\;\tilde{f}\;\;$
  2. $\tilde{g_1}\,\;\neq\;\;\tilde{g_2}\;\;$
  3. $g_1\;\;{\scriptstyle\circ}\;\;f\,\;\neq\;\;g_2\;\;{\scriptstyle\circ}\;\;f\;\;\;$ (otherwise, $f\;$ being epic would imply $g_1 = g_2$, and this would lead to a contradiction with (2))

The fact that $f\;$ is surjective makes it very difficult for me to envision a situation in which both (1) and (2) could possibly hold. (Also, if truth be told, maybe I have no business fighting with this problem, given that my familiarity with hTop doesn't go much beyond having read the definitions needed to follow the above.)

Any suggestions for suitable $(Z, \tau\,)$, $g_1$, and $g_2$ would be much appreciated.

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    Thanks! I will ask again if you use it again :)2012-01-06

1 Answers 1

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Since $\mathbb{R}$ is contractible, it is homotopy equivalent to a point, hence all maps from $\mathbb{R}$ to a path-connected space are homotopy equivalent.

Now $g_1: z \mapsto z$ and $g_2: z \mapsto z^2$ aren't homotopy equivalent as maps $S^1 \to S^1$ (they are distinguished by the degree — which is a homotopy invariant; or if you prefer $g_1$ induces the identity on $\pi_1(S^1) \cong \mathbb{Z}$ while $g_2$ induces multiplication by $2$) but their composition with any $f: \mathbb{R} \to S^1$ is null-homotopic by the first paragraph.