Prove the cylinder $S^1 \times I$ is homotopy equivalent to the circle $S^1$

Question

this is a topology question:

Prove that the cylinder $S^1 \times I$ is homotopy equivalent to the circle $S^1$

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What I have:

Define $$f: S^1 \times I \rightarrow S^1, \text{with}\, (x,t)\mapsto x$$

Define $$g: S^1 \rightarrow S^1 \times I, \text{with} \, x \mapsto (x,0)$$

Then $f \circ g: S^1 \rightarrow S^1$ as $f \circ g(x) = x = id_{S^1}$, $g \circ f: S^1 \times I \rightarrow S^1 \times I$ as $g \circ f(x) ((x,t)) = g(x) = (x,0)$.

Now define $$F: (S^1 \, \text{x} \, I) \, \text{x} \, I --> S^1 \, \text{x} \, I, \, \text{with} \, ((x,t), S)--> (x, (1-s)t).$$

$F$ is a continuous map and $F((x,t),0) = (x,(1-0)t) = (x,t)= id_{S^1}(x,t)$, $F((x,t),1) = (x,(1-1)t) = (x,0) = g \circ f (x,t)$. Thus $F((x,t),0) = id_{S^1}(x,t)$ and $F((x,t),1) = g \circ f (x,t)$. So $F$ is a homotopy of $id_{S^1}$(x,t) and $g \circ f (x,t)$.

$f \circ g = id_{S^1}$ and $g \circ f \simeq id_{S^1 x I}$ , and from the definition I conclude $S^1 \times I$ and $S^1$ are homotopy equivalence.

Am I right? If not, could you help me with how to work this out?

Thank you in advance for the help!

Answer 1

Yes, this is correct. In fact, the argument generalizes to the proof that deformation retracts are homotopy equivalences.