Consider the following sequence
Where $\gamma: n \mapsto 2n $ and $\varphi: n \mapsto \overline{n}$. When using the properties of tensors we get the sequence of $\mathbb{Z}/2\mathbb{Z}$, which is clearly not exact. On the other hand, $\text{im}(\gamma\otimes1) = \{2k \otimes l \mid k \in \mathbb{Z}, l \in \mathbb{Z}/2\mathbb{Z} \} = \{0\otimes 0 \}$ and $\ker(\varphi\otimes 1) = \{k\otimes l \mid \varphi(k) = \overline{0} \text{ or } l = \overline{0} \} = \{0\otimes 0\}$. Therefore, $\text{im}(\gamma\otimes1) = \ker(\varphi\otimes 1)$, which gives us that the sequence is exact. Is my image or kernel wrong?
The sequence is indeed exact in the middle (though your method of checking this really requires more justification--in general, not every element of a tensor product $A\otimes B$ has the form $a\otimes b$, and moreover $a\otimes b$ can be $0$ even if neither $a$ nor $b$ is $0$). The place it fails to be exact is on the left: $$0\to\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}\stackrel{\gamma\otimes1}{\to}\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}$$ is not exact because the kernel of $\gamma\otimes 1$ is nontrivial.