Injection of the mapping cone of $z^2$ We define the mapping cone of $f:S^1\to S^1=:Y$, $f (z)=z^2$ as the quotient space of $S^1\times [0,1]\sqcup Y$ where $(z,0)$ and $(z',0)$ are identified and where $(z,1)$ and $f(z)$ are identified for all $z,z'\in S^1$. We call this space $C_f$.
Probably my question is very easy:
What is the injection $Y\to C_f$, so that we can consider $Y\subseteq C_f$?
Further question: How does the injection $S^1\to C_f$ look like?
An explizit formula would be helpful. Thanks!

The question is not that trivial. Though it's pretty obvious that the composition
$$Y\hookrightarrow S^1\times[0,1]\sqcup Y\xrightarrow q C_f$$ is injective, you still need to show that it's a homeomorphism onto its image, i.e. an embedding. To this end, let $C\subseteq Y$ be closed. Then $\bar C=f^{-1}(C)\times\{0\}\sqcup C$ is closed in $S^1\times[0,1]\sqcup Y$, and $q(\bar C)$ is closed in $C_f$ since $\bar C$ is its preimage. But $q(\bar C)=C$, so $C$ is closed in $C_f$, and this shows that $q:Y\to C_f$ is even a closed embedding.
