Continuous bijection such that induces homomorphism is no bijection

Give an example of a continuous bijection $$\phi: X \rightarrow Y$$ between path-connected spaces such that the induced homomorphism $$\phi_{*}: \pi_1(X, x) \rightarrow \pi_1(Y, \phi(x))$$ on fundamental groups is not a bijection.

Attempt: I was thinking of taking a map $$S^1 \times \mathbb{R} \rightarrow S^1 \times S^1$$ but I'm not sure how to get a bijection, such that the induced homomorphism is not a bijection. Any help is appreciated.

Let $$X=[0,2\pi)$$ and $$Y=S^1$$ and let $$p:\mathbb{R}\to S^1$$ be the quotient map. Then, $$p|_X$$ is continuous and bijective, but $$X$$ is contractible while $$Y$$ has a nontrivial fundamental group.
• What is $\pi$ precisely? What are the equivalence classes? – Kamil Jan 6 at 12:57
• It's $\theta\mapsto \exp(i\theta)$ if you want a formula. – WoolierThanThou Jan 6 at 13:01
• you might want to use a symbol other than $\pi$ for your map since your answer also contains the number $\pi$ – William Jan 6 at 14:16