G is homeomorphic to $G/H\times H$

Question

Let G be a topological group and $H\le G$.

Let $\pi: G\to G/H$ be the canonical projection and a continuous $\sigma: G/H\to G$ such that $\pi \circ \sigma = Id$.

Prove that G is homeomorphic to $G/H\times H$.

I am confused by this question because $\pi$ is a bijective morphism so we should rather have $G\cong G/H$.

Thank you for your help.

Edit

Thanks to your comments/answers I cleared my confusion. Now if I go back to the original question. Wy is $G$ homeomorphic to $H\times G/H$.

Factorization theorem gives an isomorphism $G\cong H\times G/H$ because $H=\ker \pi$ and $G/H=\text{Image}(\pi)$.

I don't see how to build this homeomorphism by hand.

Thanks again.

Answer 1

If we take $H=\mathbb{Z}$ in $G=(\mathbb{R},+)$ then $G/{H} \simeq S^1$, the unit circle, and $\mathbb{R} \not \simeq S^1 \times \mathbb{Z}$, for connectedness reasons alone...

As to your question with a continuous section, this paper looks relevant (though it's only for Abelian topological groups).

Answer 2

Hint: Consider $f:G/H\times H\rightarrow G$ defined by $f(x,y)=\sigma(x)y$ and $h:G\rightarrow G/H\times H$ defined by $h(z)=(\pi(z), (\sigma(\pi(z))^{-1}z)$ and show that $f$ is the inverse of $h$.