Is the Pontrjagin-Thom-map null-homotopic if the normal bundle $N_i$ is trivial?

Let $i\colon X\to Y$ be an embedding of two smooth and compact manifolds (without boundary) and let $N_iX$ be the normal bundle of this embedding. A Pontrjagin-Thom construction is a map $$c_i\colon Y \to Th(N_i)$$ (which is only unique up to homotopy as one chooses a tubular neighborhood (see this ncatlab page for details)).

Is $c_i$ necessarily null-homotopic, if the normal bundle $N_i$ is trivial?

No, look at $S^1 \subset S^2$. The map is $S^2 \to S^2/(S^2-S^1\times (-1,1))\stackrel{\simeq}{\to} S^2\vee S^1$ which has non trivial degree (check the induced morphism on second cohomology).
• Thank you for the answer, Dan. I think that the Thom space is homotopy equivalent to $S^2\vee S^1$ in your example and that the second map is $$S^2\vee S^1\xrightarrow{id\vee pt}S^2\vee pt\sim S^2,$$ right? Could you say a word why this composition has non trivial degree please? Thank you! – user8463524 Mar 20 '15 at 15:31
• Dear @Chris Gerig, your editing suggests that the Thom space is in fact homotopy equivalent to $S^2\vee S^1$ but how can a map from $S^2$ to $S^2\vee S^1$ have a degree? – user8463524 Mar 20 '15 at 15:33