# In search of proof that $\widetilde{e^{ix}}:\mathbb{R}\to S^1$ is not epic in $\mathbf{hTop}$

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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Dear kjo, t.b.'s nice answer captures the point completely, but it might help to add the remark that the surjectivity of a map is typically not a homotopy invariant notion (if you imagine some $X$ mapping to some $Y$, and then deforming this map, there is no reason to think it will stay surjective in general), and so perhaps it should come as no surprise that surjective maps don't have any particularly significant properties in the homotopy category. Regards, – Matt E Jan 5 '12 at 13:11
What does ISO mean? – Mariano Suárez-Alvarez Jan 5 '12 at 23:59
@MarianoSuárez-Alvarez: ISO = "in search of" – kjo Jan 6 '12 at 0:42
Dear kjo, if you think adding a "in sarch of" to a title is sufficiently useful, please do add it in full :) – Mariano Suárez-Alvarez Jan 6 '12 at 1:12
Thanks! I will ask again if you use it again :) – Mariano Suárez-Alvarez Jan 6 '12 at 3:08

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.