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Suppose $X$ is a path-connected space, and we attach a 1-cell to it with some attaching map $f : \{0,1\} \to X$ and call the resulting space $Y$.

Is $Y$ homotopy equivalent to $X \vee \mathbb{S}^1$?.

My idea was the following: let $g : [0,1] \to X$ be a path in $X$ with endpoints $g(0) = f(0)$ and $g(1) = f(1)$. Then we can stretch out the image of $g$ by attaching a strip $[0,1] \times [0,1]$ to $X$ with attaching map $h : [0,1] \times \{0\} \to X$ defined as $h(t,0) = g(t)$, the resulting space $Z$ is then homotopy equivalent to $Y$ because we can deformation retract $Z$ onto $Y$ by pushing down this added strip.

We can then push the endpoints of the attached 1-cell to the top of this strip and then squeeze the top together, and then deformation retract the squeezed strip back to the image of $g$. The resulting space is then $X \vee \mathbb{S}^1$. (See this picture for the steps visualized.)

Each step is a homotopy equivalence so this would imply that $Y$ and $X \vee \mathbb{S}^1$ are homotopy equivalent. Is this proof correct?

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Your proof is correct, but it is a particular case of more general tools we can use to show homotopy equivalence (see page 13 of Hatcher).

The basic idea is that any homotopy of an attaching map gives a homotopy equivalence between the resulting spaces.

In this case, your the path between the two points $f(0)$ and $f(1)$ gives (with a bit of work) a homotopy between between the initial attaching map, and the constant one. The constant attaching map corresponds to the wedge $X \vee S^1$, so they are homotopy equivalent via the proposition in Hatcher.

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  • $\begingroup$ The result in Hatcher that you quote requires $X$ to be a CW complex. The OP's proof does not impose that restriction. $\endgroup$
    – Rob Arthan
    Mar 24 '20 at 23:52

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