# 1-cell on path connected space is homotopy equivalent to wedge

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?

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