# Fundamental group homomorphism induced by a continuous map

**Stub grade: A**

## Definition

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces^{[Note 1]}, let [ilmath]p\in X[/ilmath] be some fixed point (to act as the base point for the fundamental group, [ilmath]\pi_1(X,p)[/ilmath]) and let [ilmath]\varphi:X\rightarrow Y[/ilmath] be a continuous map. Then^{[1]}:

- [ilmath]\varphi[/ilmath] induces a group homomorphism on the fundamental groups, [ilmath]\pi_1(X,p)[/ilmath] to [ilmath]\pi_1(Y,\varphi(p))[/ilmath], which we denote:
- [ilmath]\varphi_*:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))[/ilmath] defined as:
- [ilmath]\varphi_*:[f]\mapsto [\varphi\circ f][/ilmath]
^{[Note 2]}

For more details on the formalities of the definition see the proof: the proof, which names everything involved during the statement.

## Immediate results

### The induced homomorphism of a composition is the same as the composition of induced homomorphisms

Let [ilmath](X,\mathcal{ J })[/ilmath], [ilmath](Y,\mathcal{ K })[/ilmath] and [ilmath](Z,\mathcal{ H })[/ilmath] be topological spaces, let [ilmath]p\in X[/ilmath] be any fixed point (to act as a base point for the fundamental group [ilmath]\pi_1(X,p)[/ilmath]) and let [ilmath]\varphi:X\rightarrow Y[/ilmath] and [ilmath]\psi:Y\rightarrow Z[/ilmath] be continuous maps. Then^{[1]}:

- [ilmath](\psi\circ\varphi)_*\eq(\psi_*\circ\varphi_*)[/ilmath]
- where [ilmath]\varphi_*[/ilmath] denotes the
**fundamental group homomorphism, [ilmath]\varphi_*:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))[/ilmath], induced by [ilmath]\varphi[/ilmath]**- and "" for the others

- where [ilmath]\varphi_*[/ilmath] denotes the

Note that both of these maps have the form [ilmath]\big(:\pi_1(X,p)\rightarrow\pi_1(Z,\psi(\varphi(p))\big)[/ilmath]

### The induced homomorphism of the identity map is the identity map of the fundamental group

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space, let [ilmath]\text{Id}_X:X\rightarrow X[/ilmath] be the identity map, given by [ilmath]\text{Id}_X:x\mapsto x[/ilmath] and let [ilmath]p\in X[/ilmath] be given (this will be the basepoint of [ilmath]\pi_1(X,p)[/ilmath]) then^{[1]}:

- the
**induced map**on the fundamental group [ilmath]\pi_1(X,p)[/ilmath] is equal to the identity map on [ilmath]\pi_1(X,p)[/ilmath]- That is to say [ilmath](\text{Id}_X)_*\eq\text{Id}_{\pi_1(X,p)}:\pi_1(X,p)\rightarrow\pi_1(X,p)[/ilmath] where [ilmath]\text{Id}_{\pi_1(X,p)} [/ilmath] is given by [ilmath]\text{Id}_{\pi_1(X,p)}:[f]\mapsto [f][/ilmath]

### Homeomorphic topological spaces have isomorphic fundamental groups

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be *homeomorphic* topological spaces, let [ilmath]p\in X[/ilmath] be given (this will be the base point of the fundamental group [ilmath]\pi_1(X,p)[/ilmath]) and let [ilmath]\varphi:X\rightarrow Y[/ilmath] be that homeomorphism. Then: ^{[1]}:

- [ilmath]\pi_1(X,p)\cong\pi_1(Y,\varphi(p))[/ilmath] - where [ilmath]\cong[/ilmath] denotes group isomorphism here, but can also be used to denote topological isomorphism (AKA: homeomorphism)

That is to say:

- [ilmath]\big(X\cong_\varphi Y)\implies(\pi_1(X,p)\cong_{\varphi_*}\pi_1(Y,\varphi(p))\big)[/ilmath]

## Proof

- Refactored into own page, see: A continuous map induces a homomorphism on fundamental groups

## See also

- The induced fundamental group homomorphism of a composition of continuous maps is the same as the composition of their induced homomorphisms
- The induced fundamental group homomorphism of the identity map is the identity map of the fundamental group
- Homeomorphic topological spaces have isomorphic fundamental groups (a corollary to the previous claim)

- Types of topological retractions
- Retract - these have very useful properties
- Deformation retract

## Notes

- ↑ Of course (and as usual) there is no reason why these cannot be the same spaces
- ↑
**Caveat:**Remember that [ilmath][f][/ilmath] means [ilmath]f[/ilmath] represents the equivalence class - this definition must be "well-defined" for whichever [ilmath]f[/ilmath] we use. As such we must check that for [ilmath]\alpha\in\pi_1(X,p)[/ilmath] that for any [ilmath]f[/ilmath] and [ilmath]g[/ilmath] such that [ilmath]f,g\in\alpha[/ilmath] that [ilmath][\varphi\circ f]\eq[\varphi\circ g][/ilmath]

## References

- ↑
^{1.0}^{1.1}^{1.2}^{1.3}Introduction to Topological Manifolds - John M. Lee