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If $g:X\rightarrow Y$ is an onto continuous map, is the induced homomorphism $g_*:\pi_1(X,x)\rightarrow \pi_1(Y,y)$ onto? Also, does it matter whether $X,Y$ are path-connected or not?

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    $\begingroup$ No. Take the quotient map from the interval to the circle $\endgroup$ – leibnewtz May 10 '19 at 5:01
  • $\begingroup$ @leibnewtz Why are you answering in a comment? $\endgroup$ – Arthur May 10 '19 at 5:20
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$g_*$ can be anything in general. As noted in the comments ($[0,1]\to S^1, t\mapsto e^{2\pi i t}$), it can even be injective and not surjective. And it does not matter if the spaces are path-connected or not, as only the components of $x$ and $y$ matter for the fundamental group.

However, if there is a split of $g$, i.e. a continuous function $f\colon Y\to X$ such that $g\circ f=id_Y$, then the induced homomorphisms satisfy $g_*\circ f_*=id_{\pi_1(Y,y)}$ and as a consequence $g_*$ is surjective. By reversing the order we can make an analogous statement for injectivity.

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I think I see why you might have thought of your question.

If $g:X \to Y$ is a map of sets which is surjective, then, using the Axiom of Choice, there is a function $s: Y \to X$, such that $gs=1_Y$, called a section of $g$. However, if $g$ is a surjective continuous map of spaces, such a continuous section may not exist. If such a continuous section exists with $s(y)=x$ then the morphism $g_*$ you give is surjective.

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