# Is the induced homomorphism of an onto map necessarily onto?

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?

• No. Take the quotient map from the interval to the circle – leibnewtz May 10 '19 at 5:01
• @leibnewtz Why are you answering in a comment? – Arthur May 10 '19 at 5:20

$$g_*$$ 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.
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.